Question

A circular disc with a groove along its diameter is placed horizontally. A block of mass $1\text{kg}$ is placed as shown. The coefficient of friction between the block and all surfaces of groove in contact is $\mu =\frac{2}{5}$. The disc has an acceleration of $25\text{m}/s^2$ Find the acceleration of the block with respect to disc.

• A. $10\text{m}/s^2$
• B. $20\text{m}/s^2$
• C. $15\text{m}/s^2$
• D. $5\text{m}/s^2$

Answer 1

The correct option is A $10\text{m}/s^2$

Since the disk is a non inertial frame, a pseudo forces will act towards the right direction on the block.

Normal reaction in vertical direction $N_1=mg$
Normal reaction from side to the groove $N_2=ma\sin {37}^{\circ }$
Now, the balance the force along the groove
$m{a}_r=ma\cos {37}^{\circ }-\mu N_1-\mu N_2$ [where $N_1=mg$, $N_2=ma\sin {37}^{\circ }$ ]
Therefore, acceleration of block with respect to disc
$a_r=\frac{ma\cos {37}^{\circ }-\mu N_1-\mu N_2}{m}$
Substituting the values we get,
$a_r=\frac{1\times 25\times \cos {37}^{\circ }-\frac{2}{5}\times 1\times 10-\frac{2}{5}\times 25\times \sin {37}^{\circ }}{1}$
$a_r=10\text{m}/s^2$