Question

Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary object. The hubs and spokes are massless, so that the moment of inertia of each wheel is given by I = mr2 , where r is the radius of the wheel. Each wheel starts from rest and has a force applied tangentially at its rim. Wheel A has a mass of 1.0 kg, a diameter of 1.0 m, and an applied force of 1.0 N. Wheel B has a mass of 1.0 kg and a diameter of 2.0 m. The two wheels undergo identical angular accelerations. What is the magnitude of the force applied to wheel B?

Answer 1

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

$I=mr^2$

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

$I_A=1 \times 0.5^2\ kg.m^2$

$I_A=0.25\ kg.m^2$

T= I α

0.5= 0.25 α

$\alpha = 2\ rad/s^2$

For Wheel B

m= 1 kg

d= 2 m,r=1 m

$I_B=1 \times 1^2\ kg.m^2$

$I_B=1 \ kg.m^2$

Given that angular acceleration is same for both the wheel

$\alpha = 2\ rad/s^2$

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N