SATHEE: Laws of Motion 3 Question 17

Laws of Motion 3 Question 17

18. A circular disc with a groove along its diameter is placed horizontally.

block of mass $1 k g$ is placed as shown. The coefficient of friction between the block and all surfaces of groove $\sin θ = 3 / 5$ in contact is $μ = 2 / 5$ . 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: 18. $10 m / s^{2}$

Solution:

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 block with respect to disc

$a_{r} = \frac{m a \cos 37^{\circ} - μ N_{1} - μ N_{2}}{m}$

Substituting the values we get, $a_{r} = 10 m / s^{2}$

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Laws of Motion 3 Question 17

18. A circular disc with a groove along its diameter is placed horizontally.

Answer:

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