Question

A 16 kg block is attached to a cord that is wrapped around the rim of a wheel of radius 2 m and its hangs vertically, as shown. 1 9 16 CU 2. 3.9 4 9 m 32 The rotational inertia of the wheel is 32 kg mº. When the block is released and the cord unwinds, the magnitude of the down- ward acceleration

Answer 1

Tension in rope = T

Acceleration of block = a

Angular acceleration of flywheel = α

r = 1m

Torque on flywheel τ = T*r = 1*T

τ = I*α

T = 32α (equation 1)

Rotational acceleration (α) of flywheel and linear acceleration (a) are related by:

α = a/r = a/1 = a

Substitute for α in equation 1

T = 32*a

The 2 forces on the block are weight, (mg =16g) downwards and T upwards.

Resultant force F = mg – T

F = 16g – 32a

Applying F=ma to block:

16g – 32a = 16a

16g = 48a

a = (16/48)*g

a = 1/3*g