A 16 kg block is attached to a cord that is wrapped around the rim of a flywheel of diameter 0.40 m and hangs vertically, as shown.The rotational inertia of the flywheel is 0.50 kg∙m².When the block is released and the cord unwinds, the acceleration of the block is: A)0.15 g B)0.56 g C)0.84 g D)1.0 g E)1.3 g

Question

Quizplus · Accepted Answer

The first step may be finding the tension in the cord, which is equal to the weight of the block:

T = mg = (16 kg)(9.8 m/s^2) = 156.8 N

Next, we can use the torque equation to relate the tension to the angular acceleration of the wheel:

τ = Iα

where τ = Fr is the torque produced by the tension, F is the force component tangential to the circle, and r is the radius of the circle (half the diameter). In this case, τ = Tr = T(0.20 m).

We can also relate the angular acceleration to the linear acceleration of the block:

α = a/r

where α is in radians/s^2, a is in m/s^2, and r is in meters.

Combining these equations, we get:

T(0.20 m) = (0.50 kg∙m^2)(a/r^2)

Solving for a, we get:

a = (T(0.20 m)r^2)/(0.50 kg∙m^2)

Plugging in the values, we get:

a = [(156.8 N)(0.20 m)(0.20 m)]/(0.50 kg∙m^2)
a = 12.54 m/s^2

Finally, we can convert this to g's by dividing by the acceleration due to gravity:

a/g = 12.54 m/s^2 ÷ 9.8 m/s^2/g = 1.28

Therefore, the acceleration of the block is 1.28 g's, which is closest to choice B, 0.56 g.

A 16 Kg Block Is Attached to a Cord That