Rolling Motion - Revision history

Jfredeman6 at 13:59, 30 April 2026

2026-04-30T13:59:52Z

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	==== Comparison of Slipping vs. Non-Slipping Motion ====		==== Comparison of Slipping vs. Non-Slipping Motion ====
	'''James Fredeman – Spring 2026'''		'''James Fredeman – Spring 2026'''
	The following simulation demonstrates the dynamics of two objects—one rolling ~~without slipping~~ and one experiencing kinetic friction—as they move up and down an incline.		The following simulation demonstrates the dynamics of two objects—one slipping without rolling and one experiencing kinetic friction—as they move up and down an incline.

	View Simulation: [https://trinket.io/glowscript/24b2cde21471 ~~Interactive Rolling~~ vs. Slipping Model]		View Simulation: [https://trinket.io/glowscript/24b2cde21471 slipping vs. Slipping into Rolling Model]

	* Scenario A: Constrained No-Slip		* Scenario A: Constrained No-Slip

Jfredeman6: I am James Fredeman, and I updated the Rolling Motion wiki for Spring 2026. I added a derivation for the minimum friction needed to roll without slipping and a new section for common exam pitfalls. I also updated the computational section with a link to my GlowScript model that compares slipping and non-slipping motion. Most info is from chapter 11 from Matter & Interactions.

2026-04-28T22:03:50Z

I am James Fredeman, and I updated the Rolling Motion wiki for Spring 2026. I added a derivation for the minimum friction needed to roll without slipping and a new section for common exam pitfalls. I also updated the computational section with a link to my GlowScript model that compares slipping and non-slipping motion. Most info is from chapter 11 from Matter & Interactions.

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 ==The Main Idea==
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 These equations, along with the information given, are usually enough to solve any problem related to the acceleration of an object that rolls without slipping.
 ===A Computational Model===
 Here is a VPython simulation of a barrel rolling down a ramp without slipping. The static friction force between the barrel and the ramp affects both its translational and angular acceleration. Click "view this program" in the top left corner to view the source code.
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 [https://www.glowscript.org/#/user/YorickAndeweg/folder/PhysicsBookFolder/program/RollingMotion2 Barrel slipping down ramp simulation]
 ==Examples==
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 ::After <math>1.46 \ s</math>, the ball rolls without slipping and no longer slows down (except due to air resistance and kinetic friction as its sides rub against blades of grass, which we ignore for this problem).
 ==Connectedness==
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 *Objects that roll at a constant speed without slipping theoretically experience no friction. This is the principle behind ball bearings. Ball bearings are a layer of spheres between two sliding surfaces intended to reduce the friction between those surfaces. If they roll without slipping, there is theoretically no friction, although in reality, they often slide against each other and the casing that holds them in place.
-'''Bowling Ball'''
+'''Bowling Balls ''' James Fredeman Spring 26
-*Bowling lanes are polished to reduce friction with bowling balls, and bowling balls are very massive, resisting changes in motion. As a result, a thrown bowling ball may slide, experiencing kinetic friction, for a long time before they begin to roll without slipping. This makes bowling alleys a good place to observe the way objects go from sliding to rolling.
+*Pool balls strike one another and often begin their trajectory by sliding. Kinetic friction acts upon the balls until they transition into pure rolling motion across the felt table.
 ==History==

Laurence12799 at 01:20, 9 August 2019

2019-08-09T01:20:45Z

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	*Bowling lanes are polished to reduce friction with bowling balls, and bowling balls are very massive, resisting changes in motion. As a result, a thrown bowling ball may slide, experiencing kinetic friction, for a long time before they begin to roll without slipping. This makes bowling alleys a good place to observe the way objects go from sliding to rolling.		*Bowling lanes are polished to reduce friction with bowling balls, and bowling balls are very massive, resisting changes in motion. As a result, a thrown bowling ball may slide, experiencing kinetic friction, for a long time before they begin to roll without slipping. This makes bowling alleys a good place to observe the way objects go from sliding to rolling.

			==History==
			The importance of rolling motion became evident with the invention of the wheel 5,500 years ago. Wheels were a way to minimize the effort needed to transport objects. The wheel and many other forms of rolling motion have been in use for a long time, but it wasn't until the dawn of Classical Physics across the 1600's, 1700's, and 1800's that the motion of these commonplace objects could be given a thorough analysis quantitatively. With this, more care could be given to the design of rolling objects, applying theoretical concepts about friction and motion to make for a more effective rolling object. One of these, the Anti-lock Breaking System (ABS), is mentioned in more detail in the [[Rolling Motion#Connectedness\| Connectedness]] section.

	== See also ==		== See also ==

Laurence12799: /* Connectedness */

2019-08-09T01:12:24Z

Connectedness

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	*In cars, wheels roll along the ground as the mechanism for movement. The cars connected to the wheels are massive and do not rotate, so the rotational kinetic energy is usually a negligible fraction of the total kinetic energy of a car. Nonetheless, other aspects of rolling motion are important for car wheels. For example, it is always preferable that a car's wheels do not slip while rolling across the ground. This is because rolling without slipping is governed by static friction, and static friction force can be greater than the kinetic friction force between the same materials. This allows for more rapid acceleration and braking. The realization that car wheels should always roll without slipping led to the development of the Anti-lock Braking System (ABS), which prevents the brakes from locking up the wheels, allowing the vehicle to come to a faster stop.		*In cars, wheels roll along the ground as the mechanism for movement. The cars connected to the wheels are massive and do not rotate, so the rotational kinetic energy is usually a negligible fraction of the total kinetic energy of a car. Nonetheless, other aspects of rolling motion are important for car wheels. For example, it is always preferable that a car's wheels do not slip while rolling across the ground. This is because rolling without slipping is governed by static friction, and static friction force can be greater than the kinetic friction force between the same materials. This allows for more rapid acceleration and braking. The realization that car wheels should always roll without slipping led to the development of the Anti-lock Braking System (ABS), which prevents the brakes from locking up the wheels, allowing the vehicle to come to a faster stop.

	~~===~~Ball ~~bearings===~~		'''Ball Bearings'''

	Objects that roll at a constant speed without slipping theoretically experience no friction. This is the principle behind ball bearings. Ball bearings are a layer of spheres between two sliding surfaces intended to reduce the friction between those surfaces. If they roll without slipping, there is theoretically no friction, although in reality, they often slide against each other and the casing that holds them in place.		*Objects that roll at a constant speed without slipping theoretically experience no friction. This is the principle behind ball bearings. Ball bearings are a layer of spheres between two sliding surfaces intended to reduce the friction between those surfaces. If they roll without slipping, there is theoretically no friction, although in reality, they often slide against each other and the casing that holds them in place.

	~~===~~Bowling ~~ball===~~		'''Bowling Ball'''

	Bowling lanes are polished to reduce friction with bowling balls, and bowling balls are very massive, resisting changes in motion. As a result, a thrown bowling ball may slide, experiencing kinetic friction, for a long time before they begin to roll without slipping. This makes bowling alleys a good place to observe the way objects go from sliding to rolling.		*Bowling lanes are polished to reduce friction with bowling balls, and bowling balls are very massive, resisting changes in motion. As a result, a thrown bowling ball may slide, experiencing kinetic friction, for a long time before they begin to roll without slipping. This makes bowling alleys a good place to observe the way objects go from sliding to rolling.

	== See also ==		== See also ==

Laurence12799: /* Vehicle wheels */

2019-08-09T01:11:02Z

Vehicle wheels

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	Rolling motion can be observed in several parts of everyday life, including in vehicle wheels, in industrial applications, and in a variety of sports and games. Below are a few examples of scenarios in which rolling motion plays a critical role.		Rolling motion can be observed in several parts of everyday life, including in vehicle wheels, in industrial applications, and in a variety of sports and games. Below are a few examples of scenarios in which rolling motion plays a critical role.

	~~===~~Vehicle ~~wheels===~~		'''Vehicle Wheels'''

	In cars, wheels roll along the ground as the mechanism for movement. The cars connected to the wheels are massive and do not rotate, so the rotational kinetic energy is usually a negligible fraction of the total kinetic energy of a car. Nonetheless, other aspects of rolling motion are important for car wheels. For example, it is always preferable that a car's wheels do not slip while rolling across the ground. This is because rolling without slipping is governed by static friction, and static friction force can be greater than the kinetic friction force between the same materials. This allows for more rapid acceleration and braking. The realization that car wheels should always roll without slipping led to the development of the Anti-lock Braking System (ABS), which prevents the brakes from locking up the wheels, allowing the vehicle to come to a faster stop.		*In cars, wheels roll along the ground as the mechanism for movement. The cars connected to the wheels are massive and do not rotate, so the rotational kinetic energy is usually a negligible fraction of the total kinetic energy of a car. Nonetheless, other aspects of rolling motion are important for car wheels. For example, it is always preferable that a car's wheels do not slip while rolling across the ground. This is because rolling without slipping is governed by static friction, and static friction force can be greater than the kinetic friction force between the same materials. This allows for more rapid acceleration and braking. The realization that car wheels should always roll without slipping led to the development of the Anti-lock Braking System (ABS), which prevents the brakes from locking up the wheels, allowing the vehicle to come to a faster stop.

	===Ball bearings===		===Ball bearings===

Laurence12799: /* 4. (Difficult) */

2019-08-09T01:10:09Z

4. (Difficult)

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	:::<math>a = \frac{2}{3}g\sin\theta</math>		:::<math>a = \frac{2}{3}g\sin\theta</math>

	===~~4. (~~Difficult)===		===Difficult===

	A golf player putts a .~~046kg~~ golf ball with radius of .~~043m~~. It slides across the green, initially without rotating, with an initial speed of 6m/s. The coefficient of kinetic friction between the ball and the turf is .12~~. How long does it take for the ball to stop slipping? Assume that the golf ball is a perfect sphere that has its mass uniformly distributed throughout~~.		A golf player putts a <math>0.046 \ kg</math> golf ball with a radius of <math>0.043 \ m</math>. It slides across the green, initially without rotating, with an initial speed of <math>6 \ \frac{m}{s}</math>. The coefficient of kinetic friction between the ball and the turf is <math>0.12</math>.

	~~Solution~~:		:'''a) How long does it take for the ball to stop slipping? Assume the golf ball is a perfect sphere that has its mass uniformly distributed throughout.'''

	The magnitude of the friction force is given by		::The magnitude of the friction force is given by:

	<math>F_k = \mu_k * N = .12 * .046 * 9.8 = .054</math>N		:::<math>F_k = \mu_k N = 0.012 \times 0.046 \times 9.8 = 0.054 \ \text{N}</math>

	This friction force has two effects. Firstly, it reduces the ball's translational speed. Secondly, it applies a torque to the ball, causing its angular speed to increase. The ball stops slipping when the ball's angular speed reaches a value corresponding to its translational speed so that the bottom-most point on the ball does not move relative to the ground.		::This friction force has two effects. Firstly, it reduces the ball's translational speed. Secondly, it applies a torque to the ball, causing its angular speed to increase. The ball stops slipping when the ball's angular speed reaches a value corresponding to its translational speed so that the bottom-most point on the ball does not move relative to the ground.

	First, let us find the translational speed of the ball as a function of time:		::First, let us find the translational speed of the ball as a function of time:

	<math>v_f = v_0 + at</math>		:::<math>v_f = v_0 + at</math>

	<math>v_0 = 6~~</math>~~m/s~~, <math>~~a = \frac{f}{m} = - \frac{.054}{.046}~~</math>~~m/s~~<sup>~~2</~~sup~~>		:::<math>v_0 = 6 \ \frac{m}{s} \quad \And \quad a = \frac{F_{k}}{m} = - \frac{0.054}{0.046} = -1.17 \ \frac{m}{s^2}</math>

	<math>v(t) = 6 - 1.17t</math>		:::<math>v(t) = 6 - 1.17t</math>

	Next, let us find the angular speed of the ball as a function of time:		::Next, let us find the angular speed of the ball as a function of time:

	<math>\omega_f = \omega_0 + \alpha t</math>		:::<math>\omega_f = \omega_0 + \alpha t</math>

	<math>\omega_0 = 0~~</math>, <math>~~\alpha = \frac{\tau}{I} = \frac{.054 * .043}{\frac{2}{5} * .046 * .043^2} = 68.25~~</math>~~s~~<math>~~^{-2}</math>		:::<math>\omega_0 = 0 \quad \And \quad \alpha = \frac{\tau}{I} = \frac{F_{k}r}{I} = \frac{0.054 \times 0.043}{\frac{2}{5} \times 0.046 \times 0.043^2} = 68.25 \ \frac{rads}{s^{2}}</math>

	<math>\omega (t) = 68.25t</math>		:::<math>\omega (t) = 68.25t</math>

	When the ball no longer slips, <math>v = r \omega</math>.		::When the ball no longer slips, <math>v = r \omega</math>:

	<math>6 - 1.17t = .043 * 68.25t</math>		:::<math>6 - 1.17t = 0.043 \times 68.25t</math>

	<math>~~6 =~~ 4.10t</math>		:::<math>4.10t = 6</math>

	<math>t = 1.46</math>s.		:::<math>t = 1.46 \ s</math>.

	After 1.~~46s~~, the ball rolls without slipping and no longer slows down (except due to air resistance and kinetic friction as its sides rub against blades of grass, which we ignore for this problem).		::After <math>1.46 \ s</math>, the ball rolls without slipping and no longer slows down (except due to air resistance and kinetic friction as its sides rub against blades of grass, which we ignore for this problem).

	==Connectedness==		==Connectedness==

Laurence12799: /* (Middling */

2019-08-09T00:53:54Z

(Middling

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	:::<math>h_{final} = \frac{7v^2}{10g}</math>		:::<math>h_{final} = \frac{7v^2}{10g}</math>

	===(Middling===		===Middling===

	A uniform cylinder of mass <math>M</math> and radius <math>R</math> is placed on a ramp inclined at an angle <math>\theta</math> above the horizontal. It rolls down without slipping.		A uniform cylinder of mass <math>M</math> and radius <math>R</math> is placed on a ramp inclined at an angle <math>\theta</math> above the horizontal. It rolls down without slipping.

Laurence12799: /* 3. (Simple) */

2019-08-09T00:53:03Z

3. (Simple)

← Older revision		Revision as of 20:53, 8 August 2019
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	:::<math>a = \frac{2}{3}g\sin\theta</math>		:::<math>a = \frac{2}{3}g\sin\theta</math>

	~~===3. (Simple)===~~

	Consider the cylinder from the previous problem. The coefficient of static friction between the cylinder and the ramp is <math>\mu_s</math>. What is the steepest the ramp can be without the cylinder slipping?

	~~Solution:~~

	~~In the process of solving the previous problem, we found that~~

	~~<math>f_s = \frac{aM}{2}</math>~~

	~~Substituting in the value for <math>a</math> that was ultimately found yields~~

	~~<math>f_s = \frac{\frac{2}{3} g\sin\theta M}{2} = \frac{1}{3} Mg\sin\theta</math>~~

	When the ramp is the steepest it can be without the cylinder slipping, <math>f_s</math> has its maximum value, namely <math>\mu_s * N</math>. The normal force <math>N</math> is <math>Mg\cos\theta</math>, the component of the cylinder's weight perpendicular to the ramp.

	~~<math>\mu_s * Mg\cos\theta = \frac{1}{3} Mg\sin\theta</math>~~

	~~This equation can be simplified and rearranged to yield~~

	~~<math>3 \mu_s = \frac{\sin\theta}{\cos\theta}</math>~~

	~~<math>3 \mu_s = \tan\theta</math>~~

	~~<math>\theta = \tan^{-1} (3 \mu_s)</math>~~

	===4. (Difficult)===		===4. (Difficult)===

Laurence12799: /* 2. (Middling) */

2019-08-09T00:51:51Z

2. (Middling)

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	:::<math>h_{final} = \frac{7v^2}{10g}</math>		:::<math>h_{final} = \frac{7v^2}{10g}</math>

	===2. (Middling)===		===(Middling===

	A uniform cylinder of mass <math>M</math> and radius <math>R</math> is placed on a ramp inclined at an angle <math>\theta</math> above the horizontal. It rolls down without slipping. ~~What is the translational acceleration of the cylinder?~~		A uniform cylinder of mass <math>M</math> and radius <math>R</math> is placed on a ramp inclined at an angle <math>\theta</math> above the horizontal. It rolls down without slipping.

	[[File:rollingmotioncylinderramp.png]]		[[File:rollingmotioncylinderramp.png\|center]]

	~~Solution~~:		:'''a) What is the translational acceleration of the cylinder?'''

	Let us use the system of equations set out in the "Solving problems about rolling objects" section:		::Let us use the system of equations set out in the [[Rolling Motion#Solving problems about rolling objects\| Solving problems about rolling objects]] section:

	~~<ol>~~		:::<math>a = \frac{f_{ext} - f_s}{m}</math>
	~~<li>~~<math>a = \frac{f_{ext} - f_s}{m}</math~~></li~~>		:::<math>\alpha = \frac{f_s r}{I}</math>
	~~<li>~~<math>\alpha = \frac{f_s r}{I}</math~~></li~~>		:::<math>a = \alpha r</math>
	~~<li>~~<math>a = \alpha r</math~~></li>~~
	~~</ol~~>

	For this problem, <math>m</math> and <math>r</math> are given to be <math>M</math> and <math>R</math> respectively. <math>f_{ext}</math> is <math>Mg\sin\theta</math> because that is the component of gravity directed parallel to the ramp and therefore not balanced by the normal force. <math>I</math> is <math>\frac{1}{2}MR^2</math> because the object in question is a cylinder. <math>f_s</math>, <math>a</math>, and <math>\alpha</math> are the three unknowns.		::For this problem, <math>m</math> and <math>r</math> are given to be <math>M</math> and <math>R</math> respectively. <math>f_{ext}</math> is <math>Mg\sin\theta</math> because that is the component of gravity directed parallel to the ramp and therefore not balanced by the normal force. <math>I</math> is <math>\frac{1}{2}MR^2</math> because the object in question is a cylinder. <math>f_s</math>, <math>a</math>, and <math>\alpha</math> are the three unknowns.

	Making these substitutions, the system of equations is now		::Making these substitutions, the system of equations is now:

	~~<ol>~~		:::<math>a = \frac{Mg\sin\theta - f_s}{M}</math> '''(1)'''
	~~<li>~~<math>a = \frac{Mg\sin\theta - f_s}{M}</math>~~</li>~~		:::<math>\alpha = \frac{f_s R}{\frac{1}{2}MR^2} = 2 \frac{f_s}{MR}</math> '''(2)'''
	~~<li>~~<math>\alpha = \frac{f_s R}{\frac{1}{2}MR^2} = 2 \frac{f_s}{MR}</math>~~</li>~~		:::<math>a = \alpha R</math> '''(3)'''
	~~<li>~~<math>a = \alpha R</math>~~</li>~~
	~~</ol>~~

	Now let us solve the system using the familiar technique of substitution. Substituting equation 2 into equation 3 yields		::Now let us solve the system using the familiar technique of substitution. Substituting equation '''2''' into equation '''3''' yields:

	<math>a = 2 \frac{f_s}{MR} R = 2 \frac{f_s}{M}</math>		:::<math>a = 2 \left(\frac{f_s}{MR}\right) R = 2 \frac{f_s}{M}</math>

	Rearranging this equation to isolate <math>f_s</math> yields		::Rearranging this equation to isolate <math>f_s</math> yields:

	<math>f_s = \frac{aM}{2}</math>		:::<math>f_s = \frac{aM}{2}</math>

	Substituting this into equation 1 yields		::Substituting this into equation '''1''' yields:

	<math>a = \frac{Mg\sin\theta - \frac{aM}{2}}{M} = g\sin\theta - \frac{a}{2}</math>		:::<math>a = \frac{Mg\sin\theta - \frac{aM}{2}}{M} = g\sin\theta - \frac{a}{2}</math>

	Adding <math>\frac{a}{2}</math> to both sides yields		::Adding <math>\frac{a}{2}</math> to both sides yields:

	<math>\frac{3}{2}a = g\sin\theta</math>		:::<math>\frac{3}{2}a = g\sin\theta</math>

	Multiplying both sides by <math>\frac{2}{3}</math> yields		::Multiplying both sides by <math>\frac{2}{3}</math> yields:

	<math>a = \frac{2}{3}g\sin\theta</math>		:::<math>a = \frac{2}{3}g\sin\theta</math>

	===3. (Simple)===		===3. (Simple)===

Laurence12799: /* 1. (Simple) */

2019-08-09T00:42:26Z

1. (Simple)

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	==Examples==		==Examples==

	===~~1. (~~Simple)===		===Simple===

	A sphere has mass <math>M</math> and radius <math>R</math>. It is rolling without slipping at a speed <math>v</math>. It comes to the foot of a ramp inclined at an angle <math>\theta</math> above the horizontal and begins to roll up it. ~~What vertical height does the sphere reach relative to its initial height?~~		A sphere has a mass <math>M</math> and radius <math>R</math>. It is rolling without slipping at a speed <math>v</math>. It comes to the foot of a ramp inclined at an angle <math>\theta</math> above the horizontal and begins to roll up it.

	[[File:rollingmotionupramp.png]]		[[File:rollingmotionupramp.png\|center]]

	~~Solution~~:		:'''a) What vertical height does the sphere reach relative to its initial height?'''

	Let us use an energy approach since the problem asks for a change in height, which directly translates to a change in gravitational potential energy, and since no information about time is necessary.		::Let us use an energy approach since the problem asks for a change in height, which directly translates to a change in gravitational potential energy, and since no information about time is necessary.

	Let us find the sphere's initial kinetic energy. When the sphere is first rolling, it has translational kinetic energy <math>\frac{1}{2}Mv^2</math> and rotational kinetic energy <math>\frac{1}{2}I\omega^2</math>.		::Let us find the sphere's initial kinetic energy. When the sphere is first rolling, it has translational kinetic energy <math>\left(\frac{1}{2}Mv^2\right)</math> and rotational kinetic energy <math>\left(\frac{1}{2}I\omega^2\right)</math>.

	Since the sphere rolls without slipping, we know that <math>\omega = \frac{v}{R}</math>.		::Since the sphere rolls without slipping, we know that <math>\omega = \frac{v}{R}</math>.

	Since the rolling object is a sphere, we know that <math>I = \frac{2}{5}MR^2</math>.		::Since the rolling object is a sphere, we know that <math>I = \frac{2}{5}MR^2</math>.

	Substituting these into the rotational kinetic energy formula yields a rotational kinetic energy of <math>\frac{1}{2}I\omega^2 = \frac{1}{2} \frac{2}{5}MR^2 (\frac{v}{R})^2 = \frac{1}{5}Mv^2</math>		::Substituting these into the rotational kinetic energy formula yields a rotational kinetic energy of <math>\frac{1}{2}I\omega^2 = \frac{1}{2} \frac{2}{5}MR^2 \left(\frac{v}{R}\right)^2 = \frac{1}{5}Mv^2</math>

	~~<math>KE_{tot} = KE_{rot} + KE_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{7}{10}Mv^2</math>~~		::Therefore, the initial kinetic energy of the rolling sphere is:

	~~When the sphere reaches its highest point along the ramp, it is neither moving translationally nor rotating, so all of the above energy has been converted to gravitational potential energy.~~		:::<math>KE_{tot} = KE_{rot} + KE_{trans} = \frac{1}{5}Mv^2 + \frac{1}{2}Mv^2 = \frac{7}{10}Mv^2</math>

	~~<math>Mgh_{final} = KE_{tot~~, ~~initial} = \frac{7}{10}Mv^2</math>~~		::When the sphere reaches its highest point along the ramp, it is neither moving translationally nor rotating, so all of the above energy has been converted to gravitational potential energy.

	<math>h_{final} = \frac{7v^2}{10g}</math>		:::<math>Mgh_{f} = KE_{tot, initial} = \frac{7}{10}Mv^2</math>

			::Solving for the final height gives:

			:::<math>h_{final} = \frac{7v^2}{10g}</math>

	===2. (Middling)===		===2. (Middling)===