Laws Of Motion Ques 25
- A circular disc with a groove along its diameter is placed horizontally.
block of mass $1 kg$ is placed as shown. The coefficient of friction between the block and all surfaces of groove $\sin \theta=3 / 5$ in contact is $\mu=2 / 5$. The disc has an acceleration of $25 m / s^{2}$. Find the acceleration of the block with respect to disc.
$(2006,6\ M)$
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Answer:
Correct Answer: 25.$10 m / s^{2}$
Solution:
Formula:
- Normal reaction in vertical direction $N _1=m g$
Normal reaction from side to the groove $N_2 = m a \sin 37^{\circ}$
Therefore, acceleration of the block with respect to the disc
$$ a _r=\frac{m a \cos 37^{\circ}-\mu N _1-\mu N _2}{m} $$
Substituting the values we get, $a _r=10 m / s^{2}$
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