Question

A circular disc with a groove along its diameter is placed horizontally. A ball of mass $1\text{kg}$ is placed in it as shown. The co-efficient of friction between the ball and all surfaces of the groove in contact is $\mu =\frac{2}{5}$. The disc has an acceleration of $25{\text{m/s}}^2$. Then, the acceleration of the ball with respect to disc will be
[Take $\theta ={37}^{\circ }$ and $g=10{\text{m/s}}^2$]

• A. $10{\text{m/s}}^2$
• B. $12{\text{m/s}}^2$
• C. $14{\text{m/s}}^2$
• D. $16{\text{m/s}}^2$

Answer 1

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

FBD of the ball in disc frame: (horizontal plane)


Here, $25\text{N}$ is the pseudo force $(F_P=ma)$

$f$ is friction due to side surface of the groove.

$f^{\prime }$ is friction due to bottom surface of the groove.

$N$ is normal reaction due to side surface of the groove

Along x - direction,
$N=25\sin \theta =15\text{N}$ -----(1)

Along y - direction,
$F_{net}=25\cos \theta -(f+f^{\prime })$ ---- (2)

$\Rightarrow ma=20-(\mu N+\mu mg)$

where $a$ is the acceleration of the ball along the groove i.e w.r.t the disc.
$\Rightarrow 1\times a=20-(\frac{2}{5}\times 15+\frac{2}{5}\times 1\times 10)$

$\Rightarrow a=20-(6+4)\Rightarrow a=10{\text{m/s}}^2$

Hence, option (a) is the correct answer.