For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is:
A)1
B)2
C)1/2
D)4
E)1/4

Question

Quizplus · Accepted Answer

Let ω be the angular velocity, α be the angular acceleration, v1 be the velocity of a point on the rim, and v2 be the velocity of a point halfway between the center and the rim. Then we have:

v1 = ωR
v2 = ω(R/2)

The tangential acceleration of a point on the rim is a1 = Rα, and the tangential acceleration of a point halfway between the center and the rim is a2 = (R/2)α.

By the formula for tangential acceleration, we have

a1 = Rα = dv1/dt = R(dω/dt)
a2 = (R/2)α = dv2/dt = (R/2)(dω/dt)

Dividing these two equations by each other,

a1/a2 = (R/R/2) = 2

But the tangential acceleration is also equal to the ratio of the speeds squared:

a1/a2 = v1^2/v2^2

So we have:

v1^2/v2^2 = 2

Taking the square root of both sides,

v1/v2 = sqrt(2)

Therefore, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is 2.

For a Wheel Spinning with Constant Angular Acceleration on an Axis