A wheel rotates about a fixed axis with a constant angular acceleration of 4.0 rad/s². The diameter of the wheel is 40 cm. What is the linear speed of a point on the rim of this wheel at an instant when that point has a total linear acceleration with a magnitude of 1.2 m/s²? A) 39 cm/s B) 42 cm/s C) 45 cm/s D) 35 cm/s E) 53 cm/s

Question

Quizplus · Accepted Answer

We can use the equations of rotational kinematics to find the angular velocity and then relate it to the linear velocity using the formula v = ωr.

We know that the angular acceleration is constant, so we can use the equation:

θ = ω0t + 1/2 αt^2

where θ is the angular displacement, ω0 is the initial angular velocity (which we can assume is 0), α is the angular acceleration, and t is the time.

Since the wheel starts from rest, we can simplify this equation to:

θ = 1/2 αt^2

We can solve for t:

t = sqrt(2θ/α)

The linear acceleration can be related to the tangential acceleration using the formula:

a = rα

where r is the radius of the wheel.

We can rearrange this to solve for α:

α = a/r

Substituting the given values:

α = 1.2 m/s^2 / 0.2 m = 6 rad/s^2

We can now find θ:

θ = 1/2 (6 rad/s^2) (t^2)

θ = 3t^2

Substituting for t:

t = sqrt(2θ/α) = sqrt(2(3t^2)/6) = t

Solving for t, we get:

t = sqrt(1/2) s

Now we can find the angular velocity using:

ω = ω0 + αt

ω = (0) + (6 rad/s^2)(sqrt(1/2) s)

ω = 3 sqrt(2) rad/s

Finally, we can find the linear velocity using:

v = ωr

v = (3 sqrt(2) rad/s)(0.2 m)

v = 0.6 sqrt(2) m/s

Converting to cm/s and rounding to the nearest integer, we get:

v = 42 cm/s, which corresponds to choice (B).

A Wheel Rotates About a Fixed Axis with a Constant