Question

M, a solid cylinder (M=1.35 kg, R=0.117 m) pivots on a thin, fixed, frictionless bearing. A string wrapped around the cylinder pulls downward with a force F which equals the weight of a 0.670 kg mass, i.e., F = 6.573 N. Calculate the angular acceleration of the cylinder. Tries 0/10 If instead of the force F an actual mass m = 0.670 kg is hung from the string, find the angular acceleration of the cylinder. Tries 0/10 How far does m travel downward between 0.450 s and 0.650 s after the motion begins? Tries 0/10 The cylinder is changed to one with the same mass and radius, but a different moment of inertia. Starting from rest, the mass now moves a distance 0.470 m in a time of 0.510 s. Find Icm of the new cylinder.

Answer 1

Moment of inertia for a solid disk = I = mr^2/2

A) Sum of the moments about the center of the disk = I*alpha = F*r

Sub in I = m*r^2/2
m*r^2/2 * alpha = F*r
m*r / 2 * alpha = F

Solve for alpha
alpha = 2*F / (m*r)

Plug in numbers
alpha = 2 * 6.573 N / (1.35 kg * .117 m) = 83.25 rad/s^2

B) Same thing, becuase in the first part F = 6.573 N. With a 0.670 kg mass, the new force is weight, and weight = m*g... W = .670 kg * 9.81 m/s^2 = 6.573 N, which is the same as F so your angular acceleration will come out the same.

C) For this part you will need to find the angle it moves between in those seconds.
Equation of motion
Theta_f = Thete_i + w_i*t + 1/2*alpha*t^2

In this case, assume Theta_i = 0 rad and w_i = 0 rad/s
Theta_f = 1/2*alpha*t^2

Plug in .450 s
Theta_f = 1/2 * 83.25 rad/s^2 * (.450 s)^2 = 8.42 rad

Plug in .650 s
Theta_f = 1/2 * 83.25 rad/s^2 * (.650 s)^2 = 17.58 rad

Subtract the two
17.58 rad - 8.42 rad = 9.16 rad

So, the cylinder rotates 9.16 rads in between those seconds. Simply use the arc length formula to find the distance traveled.

L = r*theta = 0.117 m * 9.16 rad
L = 1.07 m

D) Use the distance and time and work backwards.

From the arc length formula
L = r*theta ==> theta = L/r = 0.470 m / 0.117 m = 4.01 rad

Now input into the same equation of motion and solve for alpha
Theta = 1/2*alpha*t^2

alpha = 2*theta / t^2
alpha = 2*4.01 rad / (.510 s)^2 = 30.83 rads^2

Now back to the Moment equation
Sum of the moments about the center = I*alpha = F*r

I = F*r / alpha
I = 6.573 N * 0.117 m / (30.83 rad/s^2) = 0.0249 kg*m^2