Question 269: Laws of Motion

Question

A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 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 $25 \mathrm{~m} / \mathrm{s}^2$. Find the acceleration of the block with respect to disc.

Accepted Answer

Horizontal disc has a normal reaction $N_1=m g$. The normal reaction $\mathrm{N}_2=m g \sin 	heta$. acceleration of the man with respect to the disc is $a^{\prime}$.
$$ \begin{aligned} a^{\prime} & =\frac{\mathrm{R}}{m}=a \cos 	heta-\mu_k \mathrm{~g}-\mu_k a \sin 	heta \ a^{\prime} & =25\left(\frac{4}{5}ight)-\left(\frac{2}{5}ight)(10)-\left(\frac{2}{5}ight)(25)\left(\frac{3}{5}ight) \ a^{\prime} & =20-4-6 \ a^{\prime} & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $$
OR
Method (II)
Normal reaction in vertical direction $\mathrm{N}_1=m g$
$\Rightarrow$ Normal reaction from side to the groove
$$ \mathrm{N}_2=\mathrm{ma} \sin 37^{\circ} $$
Therefore, acceleration of block with respect to disc.
$$ \begin{aligned} a r & =\frac{m a \cos 37^{\circ}-\mu \mathrm{N}_1-\mu \mathrm{N}_2}{m} \ & =\frac{m a \cos 37^{\circ}-\mu m g-\mu m a \sin 37^{\circ}}{m} \ a r & =\frac{1 	imes 25 	imes \frac{4}{5}-\frac{2}{5} 	imes 1 	imes 10-\frac{2}{5} 	imes \frac{3}{5} 	imes 1 	imes 25}{1} \ a r & =20-4-6 \ a r & =10 \mathrm{~m} / \mathrm{s}^2 . \end{aligned} $$

Laws of Motion

Integer / Numeric Answer Type