Question

A force of 30 N is applied tangentially to the rim of a solid disk of radius 0.14 m. The disk rotates about an axis through its center and perpendicular to its face with a constant angular acceleration of 115 rad/s2. Determine the mass of the disk.

Answer 1

Answer:

The mass of the disk is 3.73 kg

Explanation:

Given;

Applied force, F =  30 N

radius of solid disk, r = 0.14 m

angular acceleration of the disk, α = 115 rad/s²

Torque on the  rim of a solid disk is given as;

τ = F x r = I x α

where;

F is the applied force

r is the radius of the solid disk

I is moment of inertia

α is angular acceleration

Moment of inertia, I, of solid disk is given as;

I = ¹/₂mr²

m is the mass of the solid disk

Now, substitute this into the above equation;

F x r = I x α

F x r = ¹/₂mr² x α

F = ¹/₂mrα

m = 2F / rα

m = (2 x 30) / (0.14 x 115)

m = 3.73 kg

Thus, the mass of the disk is 3.73 kg

Answer 2

Answer:

3.8 kg

Explanation:

The torque, τ, due to the force, F, is given by

$\tau = F\times r$

where r is the radius.

This torque is also given by

$\tau = I\times\alpha$

where I and α are respectively the moment of inertia and the angular acceleration.

For a solid disk, its moment of inertia for an axis though its centre and perpendicular to its face is given by

$I = \frac{1}{2}mr^2$

where m is its mass and r is its radius.

Hence, we have

$\tau = F\times r = I\times\alpha=\frac{1}{2}mr^2\times\alpha$

$m = \dfrac{2F}{r\alpha} = \dfrac{2(30\ \text{N})}{(0.14\ \text{m})(115\ \text{rad/s}^2)} = 3.8\ \text{kg}$