Question

A 5.5Kg block is hanging from a rope that is wrapped around the outside of a 13Kg flywheel disk witha radius of 33cm that is hagning form the ceilin. Friction in the flywheel provides a constant torque of 2.5Nm. When the block is released what is the magnitude of its acceleration as it falls

Answer 1

Answer:

$3.9m/s^{2}$

Explanation:

Using second law of motion

$a =\frac {m1 * g - \frac {T}{r}}{m1 + 0.5 * m2}$ where m1 is mass of block, m2 is mass of flywheel, g is acceleration due to gravity whose value is taken as $9.81 m/s^{2}$, T is torque and r is radius

Substituting 5.5 Kg for m1, 13 Kg for m2, 0.33 m for r, 2.5 Nm for T we obtain

$a = \frac {5.5 \times 9.81 - \frac {2.5}{0.33}}{(5.5 + 0.5 \times13)}=3.9m/s^{2}$