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Question

A circular disc with a groove along its diameter is placed horizontally. A block of mass 1 kg is placed as shown. The co-efficient of friction between the block and all surfaces of groove in contact is μ=2/5 The disc has an acceleration of 25m/s2 Find the acceleration of the block with respect to disc.
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Solution

Since the disk is a non inertial frame, we have to apply pseudo forces if we want to see the motion of the block wrt the disk.
Hence, pseudo force on the block towards right direction is 1 kg multiplied by 25 m/s2 equals 25 N.

Along the diameter it will be 25cosθ = 20 N.
The normal force between the block and the mass is 25sinθ = 15 N.
Hence the frictional force will be 15x
25 = 6 N.
So the force along the axis is (20-6) = 14 N.
Hence acceleration = 14
m/s2

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A small cubical block of mass 1 kg is placed inside a rough rectangular groove μ=12 made in a circular table as shown in the figure. The table starts rotating with angular acceleration 1rads2 in a horizontal plane about its axis. The time 10Ksec after which the blocks will start motion with respect to the table. Find the value of K. (Take g=10m/s2) Assume the size of block slightly smaller than the width of the groove.


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