Question

1. A wheel is turning about an axis through its center with constant angular acceleration. Starting from rest, at t=0, the wheel turns through 8.50 revolutions in t= 10.5 s . At 10.5 s the kinetic energy of the wheel is 34.0 J. For an axis through its center, what is the moment of inertia of the wheel? Express your answer with the appropriate units.

2.While redesigning a rocket engine, you want to reduce its weight by replacing a solid spherical part with a hollow spherical shell of the same size. The parts rotate about an axis through their center. You need to make sure that the new part always has the same rotational kinetic energy as the original part had at any given rate of rotation. If the original part had mass M , what must be the mass of the new part?

3. Find the moment of inertia of a hoop (a thin-walled, hollow ring) with mass M and radius R about an axis perpendicular to the hoop's plane at an edge. Express your answer in terms of the variables M and R.

4.A simple wheel has the form of a solid cylinder of radius r with a mass m uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about the axle at a constant angular speed. The wheel rotates n full revolutions in a time interval t. What is the kinetic energy K of the rotating wheel? Express your answer in terms of m, r, n, t, and π.

5.A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis through the center and perpendicular to the plane of the square. Use the parallel-axis theorem. Express your answer in terms of the variables M and a.

Answer 1

1)

θ = 1/2αt^2

8.5 revolutions = 8.5 x 2π = 53.41 rad

53.41 = 1/2α(10.5)^2

α = 0.9688 rad/s^2

ω = αt = 0.9688(10.5) = 10.7 rad/s

KE = 1/2Iω^2

34 = 1/2*I*(10.7)^2

I = 0.5939 kg m^2

2)

Solid Sphere, I (rotational inertia) = 2/5 MR^2

Hollow Sphere, I = 2/3 MR^2

Rotational Energy equals 1/2 I*w^2

Mnew/ Mold = Msolid/Mhollow = (2/3M)/(1/2M) = 5/3.

In order for the mass to be the same, you would have to multiply by 3/5.

Therefore, Ans is 3/5 or 0.6

3)

I = Io + M*R^2

I = M*R^2 + M*R^2

I = 2*MR^2

4)

In terms of m, r, n, t and, pi,

KE = Iω^2/2

I for a disk = mr^2/2

1 revolution = 2pi rad, so angular rate ω = 2pi*n/t rad/s.

Thus KE = mr^2/2*(2pi*n/t)^2/2

KE = mr^2pi^2n^2/t^2

KE = m(r*pi*n/t)^2

5)

first find out the moment of inertia of a one side about the axis using the parallel axis theorem

moment of inertia of a side about an axis through the Centre of Gravity and perpendicular to the rod = (1/12)mr^2

so

I' = I + md^2

I' = (1/12)ma^2 + m(a/2)^2

I' = ma^2 [ 1 + 3 ] /12

I' = 1/3*ma^2

just multiply this by 4 to get the moment of inertia of all four sides that means the moment of inertia of the system

about the centre axis.

so the answer is (4/3)ma^2

m is the mass of an one side

so m = M/4

so the answer = (4/3)(M/4)a^2

= 1/3*Ma^2