Answer:
3.8 kg
Explanation:
The torque, τ, due to the force, F, is given by
[tex]\tau = F\times r[/tex]
where r is the radius.
This torque is also given by
[tex]\tau = I\times\alpha[/tex]
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
[tex]I = \frac{1}{2}mr^2[/tex]
where m is its mass and r is its radius.
Hence, we have
[tex]\tau = F\times r = I\times\alpha=\frac{1}{2}mr^2\times\alpha[/tex]
[tex]m = \dfrac{2F}{r\alpha} = \dfrac{2(30\ \text{N})}{(0.14\ \text{m})(115\ \text{rad/s}^2)} = 3.8\ \text{kg}[/tex]
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
