Complete question is:
Two uniform solid spheres have the same mass of 1.65 kg, but one has a radius of 0.206 m and the other has a radius of 0.804 m. Each can rotate about an axis through its center. (a) What is the magnitude τ of the torque required to bring the smaller sphere from rest to an angular speed of 367 rad/s in 14.5 s? (b) What is the magnitude F of the force that must be applied tangentially at the sphere’s equator to give that torque? What are the corresponding values of (c) τ and (d) F for the larger sphere?
Answer:
A) τ = 0.709 N.m
B) F = 3.44 N
C) τ = 10.8 N.m
D) F = 13.43N
Explanation:
We are given;
Mass if each sphere = 1.65kg
Radius of the first sphere; r1 = 0.206m
Radius of second sphere; r2 = 0.804m
A) initial angular speed of smaller sphere; ω_i = 0 rad/s
Final angular speed of smaller sphere; ω_f = 367 rad/s
Time;t = 14.5 s
The constant angular acceleration is calculated from;
ω_f = ω_i + αt
367 = 0 + α(14.5)
Thus,
α = 367/14.5 = 25.31 rad/s²
The torque is given by the formula;
τ = Iα
Where τ is torque ; I is moment of inertia given as (2/5)Mr²
α is angular acceleration
Thus;
τ = (2/5)(1.65)(0.206)² x 25.31
τ = 0.709 N.m
B) The magnitude of the force that must be applied to give the torque τ is gotten from the formula;
τ = F•r•sin90°
0.709 = F x 0.206 x 1
F = 0.709/0.206
F = 3.44 N
C) Now for the larger sphere, we'll repeat the same procedure in a above. Thus;
The torque is given by the formula;
τ = Iα
Where τ is torque ; I is moment of inertia given as (2/5)Mr²
α is angular acceleration
Thus;
τ = (2/5)(1.65)(0.804)² x 25.31
τ = 10.8 N.m
D) Now for the larger sphere, we'll repeat the same procedure in b above. Thus, τ is gotten from the formula;
τ = F•r•sin90°
10.8 = F x 0.804 x 1
F = 10.8/0.804
F = 13.43N
