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Rolling Dynamics

Learn combined motion on a rigid body dynamics, rolling & static friction formula with examples. Practice radius of gyration formula, rigid body motion & rigid body dynamics, rolling friction examples.

Combined Motion on a Rigid Body

Angular Displacement, Angular Velocity and Angular Acceleration at Every Point is Same

  • Consider a rigid body rotating with angular velocity \(\omega\).
  • When a rigid body rotates, the angular velocity at every point on the body is same.
  • Similarly, the angular acceleration at every point on the body is also same.

  • Consider the following example as shown in figure.
  • If a rigid body rotates about A at an angle \(\theta\), then C also rotates about B by same angle \(\theta\).
  • So, all the points rotate by same angle at any given point of time hence angular velocity and angular acceleration of all points, are same.

  • Consider a disk rotating with angular velocity \(\omega\) about C, as shown in figure.
  • The angular velocity of disk with respect to A, B and C is also \(\omega\).

Angular Velocity of Rigid Body is Same about any Point Lying on it

e.g. Consider a rod having angular velocity \(\omega\) as shown in figure.

  • Velocity of point A = 0
  • Velocity of point B = \(\omega×x\)
  • Velocity of point C = \(\omega×2x\)

  • Now the angular velocity of C with respect to A

\(\omega_C=\dfrac{v}{r}\)

where \(v\) = velocity of point A  

\(r\)= distance between A and C

\(\omega_C=\dfrac{2\omega x}{2x}=\omega\)

\(\omega_C=\omega\)

  • Angular velocity of B with respect to A

\(\omega_B=\dfrac{v}{r}\)       

where \(v\) = velocity of point B

\(r\) = distance between A and B

\(\omega_B=\dfrac{\omega x}{x}=\omega\)

\(\omega_B=\omega\)

Conclusion

The angular velocity at every point is same about every point on the rigid body.

 

Application of \(\tau=I\alpha\)

  • Consider a situation in which a rigid body is under the influence of forces and net forces is non-zero.
  • To calculate torque about center of mass \((cm)\), only the torque due to external force needs to be considered, as torque due to pseudo force is zero about center of mass.
  • Thus torque is given as

\(\tau_{cm}=I_{cm}\alpha\)

where

\(\tau_{cm}\) = Torque about center of mass

\(I_{cm}\) = Moment of inertia about center of mass

\(\alpha\)= Angular acceleration of rigid body (we need to specify any point, as angular acceleration is same about all points)

Conclusion

Effective point of application of pseudo force is center of mass. There is no need to shift real forces to center of mass or any other point to calculate torque.

Applying Newton's Second Law of Motion

  • Point of application of force is not of utmost importance for applying Newton's Second Law for a system.

Sum of forces = \(ma_{cm}\)

  • Consider a rigid body with external forces acting on it.

  • Since in equation \(F=m\,a_{cm}\), only the motion of center of mass is considered and point of application of force is not important. We consider that all the forces are acting on the center of mass, to visualize it properly as shown in figure.

Illustration Questions

Choose the incorrect statement among the following regarding the figure shown.

A Angular velocity of A with respect to B is \(\omega\), then angular velocity of C with respect to D is also \(\omega\), where A, B, C and D all lying on a rigid body

B Torque about center of mass = \(I_{cm}\alpha_{\text{any point}}\)  

C Sum of forces = \(m\,a_{\text{any point}}\)

D Sum of forces = \(m\,a_{cm}\)

×

Since, the angular velocity at every point is same about every other point on a rigid body.

Hence, option (A) is correct.

Since, torque about center of mass \(\tau=I_{cm}\times\alpha_{\text{any point}}\)

where \(\alpha\) is the angular acceleration of any point as \(\alpha\) is same for all points on a rigid body.

Hence, option (B) is correct.

For the application of Newton's Second Law, a body acts as a point object at the center of mass with all the forces at this point.

i.e., Sum of forces = \(m\,a_{cm}\)

Hence, option (C) is incorrect.

For the application of Newton's Second Law, a body acts as a point object at the center of mass with all the forces at this point.

i.e., Sum of forces = \(m\,a_{cm}\)

Hence, option (D) is correct.

Choose the incorrect statement among the following regarding the figure shown.

image
A

Angular velocity of A with respect to B is \(\omega\), then angular velocity of C with respect to D is also \(\omega\), where A, B, C and D all lying on a rigid body

.

B

Torque about center of mass = \(I_{cm}\alpha_{\text{any point}}\)

 

C

Sum of forces = \(m\,a_{\text{any point}}\)

D

Sum of forces = \(m\,a_{cm}\)

Option C is Correct

Friction on a Rolling Body

  • Consider a body rolling on a rough horizontal surface with angular velocity \(\omega\).
  • When a body rolls over a rough surface, there is no slip between body and surface.
  • This situation is similar to walking, as when we walk, there is no slip between floor and shoes.

Static Friction

  • When two surfaces do not slip over each other, then the friction  between them is static in nature.
  • Static friction is always self adjusting in nature.
  • The value of static friction varies from 0 to \(\mu N\) and it is unknown to us.

Note

  • Do not worry about the direction of static friction in case of rolling. Assume friction to be in any direction and solve it.
  • After solving, the value of friction comes with sign. If it is positive, the assumed direction is correct and if it is negative, the direction of friction is opposite to that assumed.

Illustration Questions

A ring of radius R rolling over a rough surface as shown in figure. Choose the incorrect equation among the following.

A \(F-f=m\,a_{cm}\)

B \(FR+fR=I_{cm}\,\alpha\)

C \(a_{cm}=\alpha R\)

D \(f=\mu N\)

×

By Newton's second law,

Sum of forces = \(m\,a_{cm}\)

\(\implies F-f=m\,a_{cm}\)

Hence, option (A) is correct.

 

image

Torque about center of mass due to all the forces acting on a ring is given as

\(\tau=I_{cm}\alpha\)

Torque about all the force = \(I_{cm}\alpha\)

\(\implies FR+fR=I_{cm}\alpha\)

Hence, option (B) is correct.

 

image

The acceleration of center of mass of a ring of radius R rolling with angular acceleration \(\alpha\) is given as

 \(a_{cm}=\alpha R\)

Hence, option (C) is correct.

Since, the ring is rolling on a horizontal surface, thus, the friction force acting on the ring is less than \(\mu Mg\)

\(f<\mu N\)

Hence, option (D) is incorrect.

A ring of radius R rolling over a rough surface as shown in figure. Choose the incorrect equation among the following.

image
A

\(F-f=m\,a_{cm}\)

.

B

\(FR+fR=I_{cm}\,\alpha\)

C

\(a_{cm}=\alpha R\)

D

\(f=\mu N\)

Option D is Correct

Body Rolling on a Rough Surface

  • Consider a body rolling on a rough horizontal surface with angular velocity \(\omega\).
  • When a body rolls over a rough surface, there is no slip between body and surface.
  • This situation is similar to walking as when we walk, there is no slip between floor and shoes.

Static Friction

  • When two surfaces do not slip over each other, then the friction between them is static in nature.
  • Static friction is always self adjusting in nature.
  • The value of static friction varies from 0 to \(\mu N\) and it is unknown to us.

Note

  • Do not worry about the direction of static friction in case of rolling. Assume friction to be in any direction and solve it.
  • After solving, the value of friction comes with sign. If it is positive, the assumed direction is correct and if it is negative, the direction of friction is opposite to that assumed.

Illustration Questions

A disk of radius R, is rolling over a rough horizontal surface as shown in figure. Choose the correct equation among the following regarding it.

A \(F_1+F_2=ma_{cm}\)

B \(F_1-F_2=ma_{cm}\)

C \(F_1+F_2-f_s=ma_{cm}\)

D \(F_1R+F_2R-f_sR=ma_{cm}\)

×

By Newton's Second Law,

Sum of forces = \(ma_{cm}\)

\(\implies F_1+F_2-f_s=ma_{cm}\)

Hence, option (C) is correct.

image image

A disk of radius R, is rolling over a rough horizontal surface as shown in figure. Choose the correct equation among the following regarding it.

image
A

\(F_1+F_2=ma_{cm}\)

.

B

\(F_1-F_2=ma_{cm}\)

C

\(F_1+F_2-f_s=ma_{cm}\)

D

\(F_1R+F_2R-f_sR=ma_{cm}\)

Option C is Correct

Calculation of Angular Acceleration of a Body

  • Every rigid body has a moment of inertia about its center of mass.
  • Let a rigid body be having moment of inertia about its center of mass as \(nMR^2\), where n is a number and it varies for different bodies.
  • The value of n about a certain bodies, is as follows :

\(n=1\) for ring

\(n=\dfrac{1}{2}\) for disk

\(n=\dfrac{2}{5}\) for solid sphere

\(n=\dfrac{2}{3}\) for hollow sphere

  • Consider a rigid body whose moment of inertia about center of mass is \(nMR^2\), is rolling on a rough inclined plane.

 

  • To calculate the acceleration or angular acceleration of the body following steps are to be followed :

1. Rolling Constraint 

  • Since the body is in contact with the surface of inclined plane or ground surface so, both will have the same acceleration.
  • The acceleration of the ground surface is zero.

so, \(a_{cm}-\alpha R=a_{ground \;surface}\)

\(a_{cm}-\alpha R=0\)

\(a_{cm}=\alpha R\)

 

2. FBD of center of mass of a body

From FBD

\(mg\,sin\theta-f=ma_{cm}\) ... (1)

\(N-mg\,cos\theta=0\) ... (2)

3. FBD for torque about center of mass

  • Torque about center of mass is only due to f as all the other forces are passing through them.

\(fR=I_{cm}\alpha\)

Illustration Questions

A ring of radius R, rolls over a rough inclined plane, as shown in figure. Choose the incorrect equation regarding it.

A \(a_{cm}=\alpha R\)

B \(fR=mR^2\alpha\)

C \(mg\;sin\theta-f=ma_{cm}\)

D \(f=\mu mg\;cos\theta\)

×

Using, Rolling Constraint 

Since the body is in contact with the surface of inclined plane or ground surface so, both will have the same acceleration.

The acceleration of the ground surface is zero.

is, \(a_{cm}-\alpha R=a_{ground \;surface}\)

\(a_{cm}-\alpha R=0\)

\(a_{cm}=\alpha R\)

Hence, option (A) is correct.

image

Torque about center of mass due to all the forces acting is as follows 

Torque about center of mass = \(I_{cm}\alpha\)

\(fR=I_{cm}\alpha\)

Since, moment of inertia of center of mass is 

\(I_{cm}=mR^2\)

So, \(fR=mR^2\alpha\)

Hence, option (B) is correct.

image

FBD of center of mass of a body

From FBD

\(mg\,sin\theta-f=ma_{cm}\) ... (1)

\(N-mg\,cos\theta=0\) ... (2)

Hence, option (C) is correct.

image

FBD of center of mass of a body

From FBD

\(mg\,sin\theta-f=ma_{cm}\) ... (1)

\(N-mg\,cos\theta=0\) ... (2)

Hence, option (D) is incorrect.

image

A ring of radius R, rolls over a rough inclined plane, as shown in figure. Choose the incorrect equation regarding it.

image
A

\(a_{cm}=\alpha R\)

.

B

\(fR=mR^2\alpha\)

C

\(mg\;sin\theta-f=ma_{cm}\)

D

\(f=\mu mg\;cos\theta\)

Option D is Correct

Dependence of Acceleration(acm) on Radius of Gyration

  • Every rigid body has a moment of inertia about its center of mass.
  • Let a rigid body be having moment of inertia about its center of mass as \(nmR^2\), where n is a number and it varies for different bodies.
  • The value of n about a certain bodies, is as follows 

\(n=1\) for ring

\(n=\dfrac{1}{2}\) for disk

\(n=\dfrac{2}{5}\) for solid sphere

\(n=\dfrac{2}{3}\) for hollow sphere

  • Consider a rigid body whose moment of inertia about center of mass is \(nmR^2\), is rolling on a rough inclined plane.

  • To calculate the acceleration or angular acceleration of the body following steps are to be followed :

1. Rolling Constraint 

  • Since the body is in contact with the surface of inclined plane or ground surface so, both will have the same acceleration.
  • The acceleration of the ground surface is zero.

so, \(a_{cm}-\alpha R=a_{ground \;surface}\)

\(a_{cm}-\alpha R=0\)

\(a_{cm}=\alpha R\)

2. FBD of center of mass of a body

From FBD

\(mg\,sin\theta-f=ma_{cm}\) ... (1)

\(N-mg\,cos\theta=0\) ... (2)

3. FBD for torque about center of mass

  • Torque about center of mass is only due to f as all the other forces are passing through them.

\(fR=I_{cm}\alpha\)

 

  • So, we get three equations

\(a_{cm}=\alpha R\)   ... (1)

\(mg\,sin\theta-f_s=ma_{cm}\) ... (2)

\(fR=I_{cm}\alpha\) ... (3)

So, \(f_sR=mk^2\alpha\)

\(f_s=\dfrac{mk^2}{R^2}a_{cm}\)    \(\left[\therefore \alpha=\dfrac{a_{cm}}{R}\;\;\; \text{from (1)}\right]\) ....(4)

From (1) and (3)

\(mg\;sin\theta -\dfrac{mk^2}{R^2}a_{cm}=ma_{cm}\)

\(\implies mg\;sin\theta =ma_{cm}\left(1+\dfrac{k^2}{R^2}\right)\)

\(\implies a_{cm}=\dfrac{g\;sin\theta}{\left(1+\dfrac{k^2}{R^2}\right)}\)

Let  \(n=\dfrac{k^2}{R^2}\) 

where k = radius of gyration,

 R = radius of body

So, \(a_{cm}=\dfrac{g\;sin\theta}{(1+n)}\)

  • More the value of n, less will be the acceleration.

Illustration Questions

Two solid spheres (\(S_1\) and \(S_2\)) having radii \(r_1=20\,cm\) and \(r_2=50\,cm\) respectively, and two hollow cylinders (\(C_3\) and \(C_4\)) having radii \(r_3=30\,cm\) and \(r_4=40\,cm\) respectively, are placed on a plane inclined at an angle \(\theta=30^o\) with horizontal. All of them start rolling down at the ramp at the same instant without slipping. Which among of them will reach the bottom of the hill earlier? 

A Solid sphere (\(S_1\)) and hollow cylinder (\(C_4\))

B Both solid spheres (\(S_1\)and \(S_2\))

C Both hollow cylinders (\(C_3\)and \(C_4\))

D Hollow cylinder (\(C_3\)) only

×

The relation between the acceleration of the body (a) and radius of gyration (k) when a body rolls on a rough inclined surface is given as

\(a=\dfrac{g\;sin\theta}{\left(1+\dfrac{k^2}{R^2}\right)}\) ... (1)

 

From relation it is clear that more the value of \(\dfrac{k}{R}\), less will be the acceleration and thus it will take more time to reach down the plane.

For solid sphere (\(S_1\))

Radius of gyration \(k_1=\sqrt{\dfrac{2}{5}}r_1\)

\(\dfrac{k_1^2}{r_1^2}=\dfrac{2}{5}\)

So, acceleration \(a_1=\dfrac{g\;sin\theta}{1+\left(\dfrac{2}{5}\right)}\) ...(1)

For solid sphere (\(S_2\))

Radius of gyration \(k_2=\sqrt{\dfrac{2}{5}}r_2\)

\(\dfrac{k_2^2}{r_2^2}=\dfrac{2}{5}\)

So, acceleration \(a_2=\dfrac{g\;sin\theta}{1+\left(\dfrac{2}{5}\right)}\) 

For hollow cylinder (\(C_3\))

Radius of gyration \(k_3=\sqrt{1}\;r_3\)

\(\dfrac{k_3^2}{r_3^2}=1\)

So, acceleration \(a_3=\dfrac{g\;sin\theta}{1+1}=\dfrac{g\;sin\theta}{2}\)

For hollow cylinder (\(C_4\))

Radius of gyration \(k_4=\sqrt{1}\;r_4\)

\(\dfrac{k_4^2}{r_4^2}=1\)

So, acceleration \(a_4=\dfrac{g\;sin\theta}{1+1}=\dfrac{g\;sin\theta}{2}\)

 

 

Thus, acceleration of both solid spheres is same and that of both hollow cylinder is also same.

Since, the ratio of \(\dfrac{k^2}{r^2}\) of cylinders is greater than that of spheres. So, the acceleration of spheres is larger and they will reach earlier.

Two solid spheres (\(S_1\) and \(S_2\)) having radii \(r_1=20\,cm\) and \(r_2=50\,cm\) respectively, and two hollow cylinders (\(C_3\) and \(C_4\)) having radii \(r_3=30\,cm\) and \(r_4=40\,cm\) respectively, are placed on a plane inclined at an angle \(\theta=30^o\) with horizontal. All of them start rolling down at the ramp at the same instant without slipping. Which among of them will reach the bottom of the hill earlier? 

image
A

Solid sphere (\(S_1\)) and hollow cylinder (\(C_4\))

.

B

Both solid spheres (\(S_1\)and \(S_2\))

C

Both hollow cylinders (\(C_3\)and \(C_4\))

D

Hollow cylinder (\(C_3\)) only

Option B is Correct

Friction on a Rolling Body

  • Consider a body rolling on a rough horizontal surface with angular velocity \(\omega\).
  • When a body rolls over a rough surface, there is no slip between body and surface.
  • This situation is similar to walking, as when we walk, there is no slip between floor and shoes.

Static Friction

  • When two surfaces do not slip over each other, then the friction between them is static in nature.
  • Static friction is always self adjusting in nature.
  • The value of static friction varies from 0 to \(\mu N\) and it is unknown to us.

Note

  • Do not worry about the direction of static friction in case of rolling. Assume friction to be in any direction and solve it.
  • After solving, the value of friction comes with sign. If it is positive, the assumed direction is correct and if it is negative, the direction of friction is opposite to that assumed.

Illustration Questions

A ring of radius R is placed on a rough horizontal surface, as shown in figure. A force F is applied horizontally. Determine the direction of friction.

A Towards left

B Towards right

C Zero

D Can't predict

×

A ring is rolling on a rough surface with force 'F' and friction acting on it.

image

Let us assume that friction f is acting towards left of a ring.

By Newton's second law,

\(F-f=ma\)  .. (1)

 

Applying torque at the center of mass

\(fR+FR=mR^2\alpha\) ... (2)  (Moment of inertia of ring = \(mR^2\))

Since,  \(a=\alpha R\)

\(\Rightarrow \alpha =\dfrac{a}{R}\)

So, \(fR+FR={mR^2}\left(\dfrac{a}{R}\right)\)

\(\implies fR+FR=mRa\)

\(\implies R(f+F)=(ma)R\)

\(\implies f+F=F-f\)

\(\implies 2f=0\)

\(\implies f=0\)

Since, friction is zero. Hence option (C) is correct.

A ring of radius R is placed on a rough horizontal surface, as shown in figure. A force F is applied horizontally. Determine the direction of friction.

image
A

Towards left

.

B

Towards right

C

Zero

D

Can't predict

Option C is Correct

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