Question

A circular disc of radius R rolls without slipping on a rough horizontal surface. At the instant shown its linear velocity is V, linear acceleration a, angular velocity $\omega$ and angular acceleration $\alpha$. Four points A, B, C and D lie on its circumference such that the diameter AC is vertical & BD horizontal then choose the correct options. Then choose the correct option(s) from the given four options.

• A. $V_B=\sqrt{V^2+(R\omega {)}^2}$
• B. $V_C=V+R\omega$
• C. $a_A=\sqrt{(a-R\alpha {)}^2+({\omega }^{2}R{)}^2}$
• D. $a_D=\sqrt{(a+{\omega }^{2}R{)}^2+(R\alpha {)}^2}$

Answer 1

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

The correct options are
A $a_A=\sqrt{(a-R\alpha {)}^2+({\omega }^{2}R{)}^2}$
B $a_D=\sqrt{(a+{\omega }^{2}R{)}^2+(R\alpha {)}^2}$
C $V_C=V+R\omega$
D $V_B=\sqrt{V^2+(R\omega {)}^2}$

The acceleration at each given point has two components- one in the tangential direction which is given as $R\alpha$, and the other in the radial direction having a magnitude ${\omega }^{2}R$. And the linear acceleration at each point on the sphere is a.
The components of acceleration at point A are thus $a-R\alpha$ and ${\omega }^{2}R$ giving the resultant as $\sqrt{(a-r\alpha {)}^2+(R\alpha {)}^2}$.
Similarly at point D we have the resultant as $\sqrt{(a+{\omega }^{2}R{)}^2+(R\alpha {)}^2}$
Now as for the velocities again we have two components- one in horizontal direction with magnitude v and other in the tangential direction with magnitude ${\omega }^{2}R$.
Thus for point C which is at the topmost point we have the velocity as $V+R\omega$ and for point B is $\sqrt{(V{)}^2+(R\omega {)}^2}$