Question

A particle of mass m is attached to the rim of a uniform disc of mass m and radius R the disc is rolling without slipping on a stationary horizontal surface as shown in the figure At a particular instant the particle is the top most position and center of the disc has speed $V_0$ and its angular speed is $\omega$ Choose the correct regarding the motion of the system ( disc + particle ) at that instant

• A. $V_0=\omega R$
• B. kinetic energy of the system is $\frac{11}{4}m{v}_0^2$
• C. Speed of point mass m is less than $2V_0$
• D. $|\underset{V_C}{\to }-\underset{V_A}{\to }|=|\underset{V_B}{\to }-\underset{V_D}{\to }|$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $V_0=\omega R$
B kinetic energy of the system is $\frac{11}{4}m{v}_0^2$
D $|\underset{V_C}{\to }-\underset{V_A}{\to }|=|\underset{V_B}{\to }-\underset{V_D}{\to }|$

(A) is correct, as for rolling with

out slipping velocity of point B should

be zero.

As, $V_B=V_o-wR$

$\Rightarrow 0=Vo-wR\Rightarrow wR=Vo$

(B) is correct as

kinetic energy of system = KE particle +KE Disle

KE particle = $\frac{1}{2}m(VA{)}^2=\frac{1}{2}m(2Vo{)}^2=2mV{o}^2$

$K{E}_{Disk}=\frac{1}{2}mV{o}^2+\frac{1}{2}m(2Vo{)}^2+\frac{1}{2}{\left(\frac{mR}{2}\right)}^2{w}^2$

$=\frac{1}{2}mV{o}^2+{\frac{mVO}{4}}^2=\frac{3}{4}mV{o}^2$

$K{E}_{system}=2mV{o}^2+\frac{3mV{o}^2}{4}=\frac{11mV{o}^2}{4}$

(C) is incorrect , as velocity of mass

$=V_A=2V_o$

(D) is correct , as for object

with rational and translation

motion, every point w.r.t other

points on the object rotates around

them with angular velocity 'w'

As the distance b/w points (and A

and D and B are same lie $\sqrt{2}R)$

$|\overrightarrow{V_c}-\overrightarrow{V_a}|=|\overrightarrow{V_B}-\overrightarrow{V_D}|=\sqrt{2}wR$