Informative line

Constraint And Rolling

Learn combination of translational and rotational motion, rolling motion without Slipping. Practice acceleration of a point on a body and the rolling constraint requires the motion without slipping.

Combined Motion

Representation of translational motion

  • A body that is in a pure translational motion can be represented by a particle of same mass placed at its center of mass.

Representation of rotational motion

  • Rotational motion of a body about a fixed axis has to be represented by the angular variables with respect to axis of rotation.

Combination of translational and rotational motion

  • When the axis of rotation of a body is in translational motion, the motion of the body becomes complicated.

  • For this case, the motion of the body can be represented as a combination of translational motion of center of mass and the rotation of body with respect to center of mass.

Illustration Questions

A small ball of radius r is rolling in a large hemispherical bowl of radius R. The motion of center of mass of ball and motion of ball with respect to center of mass will be respectively:

A Translational circular motion, rotational motion

B Rotational motion, Rotational motion

C Straight line translational motion, Rotational motion

D Rotational Motion, Translational motion

×

The motion of ball can be expressed as

image

Representation of motion of center of mass + Motion with respect to center of mass

image

Hence, the center of mass of ball performs translational circular motion and motion of ball with respect to center of mass is rotational. 

Motion of a point is always translational.

Hence, option (A) is correct.

A small ball of radius r is rolling in a large hemispherical bowl of radius R. The motion of center of mass of ball and motion of ball with respect to center of mass will be respectively:

image
A

Translational circular motion, rotational motion

.

B

Rotational motion, Rotational motion

C

Straight line translational motion, Rotational motion

D

Rotational Motion, Translational motion

Option A is Correct

Velocity of a Point due to Motion of Center of Mass and Motion of Body with respect to Center of Mass

  • Velocity of a point is the vector sum of \(v_{CM}\) and velocity (\(v\)) of a point rotating with respect to Center of Mass

Calculation of \(\vec v\) of points P, Q, R and S on the wheel in terms of \(v_{CM}\)\(\omega\) and R

  • The motion of wheel can be seen as :

Motion of center of mass + Motion with respect to center of mass

  • Velocity of any point is the vector sum of velocity of center of mass and velocity with respect to center of mass.
  • Velocity at point P,  vP :

\(|\vec v_P|=\sqrt {(v_{CM})^2+(\omega r)^2}\)

  • Velocity at point Q,  vQ :

\(|\vec v_Q|=v_{CM}+\omega r\)

 

  • Velocity at point R,  vR :

\(|\vec v_R|=\sqrt {(v_{CM})^2+(\omega r)^2}\)

 

  • Velocity at point S,  vS :

\(|\vec v_S|=|v_{CM}-\omega r|\)

 

Illustration Questions

Find the speed of point P on the disk if the \(v_{CM}=2\ m/sec\), \(\omega=1\ rad/sec\), \( R = 2\,m \) and \(\theta =\)\(\dfrac {\pi}{3}\)

A 2 m/sec

B 0 m/sec

C 4 m/sec

D 6 m/sec

×

Motion of disk can be expressed as :

image

Velocity of any point of the disk with respect to center of mass is perpendicular to the radius of the disk.

Velocity at point P can be expressed as 

\(\vec v_P=\vec v_{P/CM}+\vec v_{CM}\)

\(\theta=\dfrac{\pi}{3}\)

image

Velocity at point P, by Parallelogram Law

\(|v_P|=\sqrt {v^2+(\omega R)^2+2v(\omega R)\, cos\dfrac {2\pi}{3}}\)

Given : \(v=v_{CM}=2\ m/sec,\ \omega=1\ rad/sec,\ R=2\ m\)

\(|v_P|=\sqrt {(2)^2+(2)^2+2×2×2\, cos\dfrac {2\pi}{3}}\)

\(v_P=2\ m/sec\)

Find the speed of point P on the disk if the \(v_{CM}=2\ m/sec\), \(\omega=1\ rad/sec\), \( R = 2\,m \) and \(\theta =\)\(\dfrac {\pi}{3}\)

image
A

2 m/sec

.

B

0 m/sec

C

4 m/sec

D

6 m/sec

Option A is Correct

Acceleration of a point on a Body

  • Acceleration at a point on a body is the vector sum of \(a_{CM}\) and  \(a\) of point with respect to center of mass.
  • A disk is in motion, as shown in figure. To calculate acceleration at any point at time t = 0 (\(\omega=0\))

  • The motion of disk can be seen as :

Motion of center of mass + Motion with respect to center of mass

 

Acceleration at P,  \(\vec a_P\) :

\(|\vec a_P|=\sqrt {a^2+(r\alpha)^2}\)

Acceleration at Q,  \(\vec a_Q\) :

\(|\vec a_Q|=a+r\alpha\)

Acceleration at R,  \(\vec a_R\) :

\(|\vec a_R|=\sqrt {a^2+(r\alpha)^2}\)

Acceleration at S,  \(\vec a_S\) :

\(|\vec a_S|=|a-r\alpha|\)

 

 

Acceleration at point T,  \(\vec a_T\) :

\(|\vec a_T|=\sqrt {a^2+(r\alpha)^2+2ar\alpha\;cos(\pi-\theta)}\)

 

 

Illustration Questions

Find magnitude of \(\vec a\) of point P on the disk at an instant such that \(\omega = 0\). Given: \(\vec a_{CM}=\)\(2\, m/sec^2\), \(\alpha=\)\(1\, rad / sec^2\), \( R = 2\,m\) and \(\theta = \dfrac {\pi}{3}\)

A 2 m/sec2

B 0 m/sec2

C 4 m/sec2

D 1 m/sec2

×

The motion of disk can be seen as :

Motion of center of mass + Motion with respect to center of mass

image

Acceleration of any point of the disk with respect to center of mass is perpendicular to the radius of the disk.

Acceleration at point P can be expressed as 

\(\vec a_P=\vec a_{P/CM}+\vec a_{CM}\)

image

Acceleration at P, By Parallelogram Law

\(|\vec a_P|=\sqrt {a^2+(\alpha R)^2+2a(\alpha R)\;cos\dfrac{2\pi}{3}}\)

Given: \(\vec a_{cm}=\)\(2\, m/sec^2\)\(\alpha=\)\(1\, rad / sec^2\), \(R = 2\,m\)

\(|\vec a_P|=\sqrt {(2)^2+(2)^2+2(2)(2)\;cos\dfrac{2\pi}{3}}\)

\(|\vec a_P|=\)\(2\, m/sec^2\)

Find magnitude of \(\vec a\) of point P on the disk at an instant such that \(\omega = 0\). Given: \(\vec a_{CM}=\)\(2\, m/sec^2\), \(\alpha=\)\(1\, rad / sec^2\), \( R = 2\,m\) and \(\theta = \dfrac {\pi}{3}\)

image
A

2 m/sec2

.

B

0 m/sec2

C

4 m/sec2

D

1 m/sec2

Option A is Correct

Rolling

Rolling is a Combination Motion

  • Rolling is a combination of translational motion, which is same as the translational motion of center of mass and rotational motion of the body with respect to Center of Mass.

Rolling Constraint

The rolling constraint requires the motion without slipping.

  • The velocity of contact point is equal.

  • The tangential acceleration of contact point is equal.

\((a_P)_{\text {tangential}}=a_{Q}\)

Condition of No Slipping

  • The velocity of contact point is same not only at the instant shown but also for any time that follows.

 

Illustration Questions

For pure rolling, which of the following condition is correct?

A \(v_C = v_P\)

B \(v_B = v_P\)

C \(v_C =r\ {\omega}\)

D \(v_A = v_P\)

×

In pure rolling, there is no slipping at the point of contact.

\(\therefore\) Velocities at point of contact must be equal.

\(v_B = v_P\)

For pure rolling, which of the following condition is correct?

image
A

\(v_C = v_P\)

.

B

\(v_B = v_P\)

C

\(v_C =r\ {\omega}\)

D

\(v_A = v_P\)

Option B is Correct

Constraint of Rolling

  • Rolling is a combination motion in which an object rotates about an axis that is itself moving, such that there is no slipping between the surfaces in contact.
  • Consider a ball, rolling on surface B.

  • Motion of Ball A can be expressed as

Constraint of Rolling

  • The velocity and tangential acceleration of contact points of two surfaces are equal.
  • Relative velocity of the contact point is zero.

\(v_P=v-\omega R=v_B\)

Case 1 : Rolling on ground when Velocity of Ground is Zero

\(v_{\text { Ground}}=0\) [ As ground is stationary ]

In that case, \(v-r\omega=0\)

or, \(v=r\omega\)

The velocities of different points of ball are

as \(v=R\ \omega\)

Hence, point P is at rest instantaneously.

  • Relative tangential acceleration of contact point is zero.

\((a_P)_{\text { tangential}}=a-r\alpha=a_B\)

  • If relative tangential acceleration is non-zero, the relative velocity at next instant would be non-zero which would result in slipping. Hence, for pure rolling, relative tangential acceleration at is also zero.

Case 2 : Rolling on Ground when acceleration of Ground is Zero

As acceleration of ground is zero

\((a_P)_t=0\)

or, \(a-r\;\alpha=0\)

\(a=r\;\alpha\)

The acceleration of different points of ball.

As \(a=r\;\alpha\)

Illustration Questions

Choose the correct option for the relation between (\(v\) and \(\omega\)) and (\(a\) and \(\alpha\)) if the disk rolls on the slab without slipping, as shown in figure.

A \(a=r\alpha,\ v=r\omega\)

B \(a=r\alpha ,\ v=r\omega+v_B\)

C \(\alpha=ar ,\ \omega=vr\)

D \(a=a_B+r\alpha ,\ v=v_B+r\omega\)

×

Relative velocity of the contact point is zero.

\(v_P=v-\omega r=v_B\)

\(v=v_B+r\omega\)

image

Relative tangential acceleration of contact point is zero.

\((a_P)_{\text { tangential}}=a-r\alpha=a_B\)

\(a=a_B+r\alpha\)

image

Choose the correct option for the relation between (\(v\) and \(\omega\)) and (\(a\) and \(\alpha\)) if the disk rolls on the slab without slipping, as shown in figure.

image
A

\(a=r\alpha,\ v=r\omega\)

.

B

\(a=r\alpha ,\ v=r\omega+v_B\)

C

\(\alpha=ar ,\ \omega=vr\)

D

\(a=a_B+r\alpha ,\ v=v_B+r\omega\)

Option D is Correct

String Wrapped on Cylinder

  • Consider a string, wrapped on a cylinder is pulled by a force and the cylinder moves. Let the cylinder be in pure rolling motion on ground (rolling without slipping)

Relation between Acceleration of Different Points

  • Acceleration of each point on the body has two components :

  • Net acceleration at different points

  • For pure rolling

\((a_P)_{tangential}=a_{ground}\)

\(\therefore a-r\alpha=0\)

\(a=r\alpha\)

  • Net acceleration at point T

\(a_T=\sqrt {a^2+a^2+2a^2\ cos(\pi-\theta)}\)

\(=\sqrt {2a^2-2a^2\;cos\,\theta}\)

\(a_T=a\sqrt {2-2\;cos\,\theta}\)

 

 

Illustration Questions

A string is wrapped on a cylinder and is being pulled horizontally. The cylinder rolls without slipping on ground. If \(a_{cm}\) of cylinder is \(a_0\). What is the acceleration of point A?

A \(a_0\)

B \(2a_0\)

C \(3a_0\)

D Zero

×

The pure rolling motion of cylinder can be represented as sum of translational and rotational motion.

image

For pure rolling, the velocity as well as tangential acceleration of the contact point are same.

image

Velocity at different points

image

Acceleration at different points

image

Net Acceleration at R

Acceleration of A is same as acceleration of point R on cylinder.

image

A string is wrapped on a cylinder and is being pulled horizontally. The cylinder rolls without slipping on ground. If \(a_{cm}\) of cylinder is \(a_0\). What is the acceleration of point A?

image
A

\(a_0\)

.

B

\(2a_0\)

C

\(3a_0\)

D

Zero

Option B is Correct

Practice Now