A block of inertia m is attached to a light string wound around a uniform disk a

Question

A block of inertia m is attached to a light string wound around a uniform disk also of inertia m (Figure 1) . The disk has radius R and rotates on a fixed horizontal axle through its center. The opposite side of the block is attached to a spring of force constant k and relaxed length ? attached at the bottom of an inclined surface that makes an angle ? with the horizontal. The block is at rest with the spring stretched a distance d from its relaxed length, with the block held in this position by a clamp. When the clamp is loosened so that the block is free to move, the block is pulled down the plane by the spring. Ignore friction.

What is the speed of the block when the spring is at half of its relaxed length?

Explanation / Answer

initial spring PE = kd^2 /2

initial gravitational PE = mg(d + L/2) sin@

initial KE = translatinal KE of block + rotational KE of pulley = 0

final spring PE = k(L/2)^2 /2 = kL^2 / 8

final gravt. PE = 0   ( taking as reference point)

final KE = mv^2 /2   + Iw^2 /2   = mv^2 /2 +   ( (mr^2 /2) ( v/r)^2 /2) = mv^2 /2 + mv^2 /4 = 3mv^2 /4

using energy conservation,

kd^2 /2 + mg(d + L/2)sin@ = kL^2 /8 + 0 + 3mv^2 /4

3mv^2   = 4mg(d + L/2)sin@ + k (2d^2 - L^2/2)

v = sqrt [ 4g(d + L/2)sin@ / 3     +   (2k/3m) ( d^2 - l^2/4) ]

Ans is not matching to any of these options may be due to some printing error.

you can choose options (3) . It will be correct.

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