How Does a Frictionless Incline Affect Crate Acceleration and Wheel Tension?

[CaptFormal]

Homework Statement

A wheel of radius R = 30.0 cm is mounted on a frictionless horizontal axis. A massless cord is wrapped around the wheel and attached to a 4.75-kg crate that slides on a frictionless surface inclined at an angle of 25.0° with the horizontal. The crate accelerates down the incline at 2.4 m/s2.

(a) What is the tension in the cord?

(b) What is the rotational inertia of the wheel about its axis of rotation?

http://schubert.tmcc.edu/enc/66/4bc10f310fba6066d747ea495c942f9f2656106d7ddf2ff5d8b54e0cfacdd98a6596246f7119a41f9feaf3573da3a023248c9ec461bb2c9f6a44610f161f83989339c3411816f6076856c10d8af94242ce73cdf70aa0bc55.gif

Homework Equations

I = MR^2

I = Inertia
M = Mass
R = Radius

The Attempt at a Solution

I thought I could solve for part (b) using the above formula but I am not getting the correct answer. And I am not even sure how to start on part (a). Any help will be appreciated.

 

[Doc Al]

I thought I could solve for part (b) using the above formula but I am not getting the correct answer.

You're not given the mass, nor can you assume that the wheel is a uniform disk. So that won't work. First solve part (a), then use it to solve (b).

And I am not even sure how to start on part (a).

What forces act on the crate? Apply Newton's 2nd law.

 

[kerimek]

For part (a), we can use Newton's second law, F=ma, to calculate the tension in the cord. The crate is accelerating down the incline at 2.4 m/s^2, so the net force in the direction of motion is F=ma = (4.75 kg)(2.4 m/s^2) = 11.4 N. This force is equal to the tension in the cord, so the tension is 11.4 N.

For part (b), the rotational inertia of the wheel can be calculated using the formula I=MR^2. However, in this case, the mass of the wheel is not given. We can use the fact that the crate is accelerating down the incline and the wheel is not slipping to find the mass of the wheel. The acceleration of the crate is also the tangential acceleration of the wheel, so we can use the formula a=rα, where r is the radius of the wheel and α is the angular acceleration. We know the tangential acceleration (2.4 m/s^2) and the radius (30 cm = 0.3 m), so we can solve for α.

α = a/r = (2.4 m/s^2)/(0.3 m) = 8 rad/s^2

Now, we can use the formula τ=Iα, where τ is the torque and I is the rotational inertia, to find the rotational inertia of the wheel. The torque is equal to the tension in the cord multiplied by the radius of the wheel.

τ = Fr = (11.4 N)(0.3 m) = 3.42 Nm

Therefore, I = τ/α = (3.42 Nm)/(8 rad/s^2) = 0.428 kgm^2

So, the rotational inertia of the wheel about its axis of rotation is 0.428 kgm^2.