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.
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
Representation of motion of center of mass + Motion with respect to center of mass
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.
Option A is Correct
Motion of center of mass + Motion with respect to center of mass
\(|\vec v_P|=\sqrt {(v_{CM})^2+(\omega r)^2}\)
\(|\vec v_Q|=v_{CM}+\omega r\)
\(|\vec v_R|=\sqrt {(v_{CM})^2+(\omega r)^2}\)
\(|\vec v_S|=|v_{CM}-\omega r|\)
A 2 m/sec
B 0 m/sec
C 4 m/sec
D 6 m/sec
Motion of disk can be expressed as :
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
\(\vec v_P=\vec v_{P/CM}+\vec v_{CM}\)
\(\theta=\dfrac{\pi}{3}\)
Velocity at point P, by Parallelogram Law
\(|v_P|=\sqrt {v^2+(\omega R)^2+2v(\omega R)\, cos\dfrac {2\pi}{3}}\)
Given : \(v=v_{CM}=2\ m/sec,\ \omega=1\ rad/sec,\ R=2\ m\)
\(|v_P|=\sqrt {(2)^2+(2)^2+2×2×2\, cos\dfrac {2\pi}{3}}\)
\(v_P=2\ m/sec\)
Option A is Correct
\(v_P=v-\omega R=v_B\)
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.
\((a_P)_{\text { tangential}}=a-r\alpha=a_B\)
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 \(a=r\;\alpha\)
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.
\(v_P=v-\omega r=v_B\)
\(v=v_B+r\omega\)
Relative tangential acceleration of contact point is zero.
\((a_P)_{\text { tangential}}=a-r\alpha=a_B\)
\(a=a_B+r\alpha\)
Option D is Correct
\((a_P)_{tangential}=a_{ground}\)
\(\therefore a-r\alpha=0\)
\(a=r\alpha\)
\(a_T=\sqrt {a^2+a^2+2a^2\ cos(\pi-\theta)}\)
\(=\sqrt {2a^2-2a^2\;cos\,\theta}\)
\(a_T=a\sqrt {2-2\;cos\,\theta}\)
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.
For pure rolling, the velocity as well as tangential acceleration of the contact point are same.
Velocity at different points
Acceleration at different points
Net Acceleration at R
Acceleration of A is same as acceleration of point R on cylinder.
Option B is Correct
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, \(\vec a_T\) :
\(|\vec a_T|=\sqrt {a^2+(r\alpha)^2+2ar\alpha\;cos(\pi-\theta)}\)
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
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
\(\vec a_P=\vec a_{P/CM}+\vec a_{CM}\)
Acceleration at P, By Parallelogram Law
\(|\vec a_P|=\sqrt {a^2+(\alpha R)^2+2a(\alpha R)\;cos\dfrac{2\pi}{3}}\)
Given: \(\vec a_{cm}=\)\(2\, m/sec^2\), \(\alpha=\)\(1\, rad / sec^2\), \(R = 2\,m\)
\(|\vec a_P|=\sqrt {(2)^2+(2)^2+2(2)(2)\;cos\dfrac{2\pi}{3}}\)
\(|\vec a_P|=\)\(2\, m/sec^2\)
Option A is Correct
The rolling constraint requires the motion without slipping.
\((a_P)_{\text {tangential}}=a_{Q}\)
A \(v_C = v_P\)
B \(v_B = v_P\)
C \(v_C =r\ {\omega}\)
D \(v_A = v_P\)
