M, a solid cylinder (M=2.07 kg, R=0.113 m) pivots on a thin, fixed, frictionless

Question

M, a solid cylinder (M=2.07 kg, R=0.113 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 0.770 kg mass, i.e., F = 7.554 N.

d. 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 0.290 m in a time of 0.450 s. Find Icm of the new cylinder
.

Explanation / Answer

The two important observations here are:

1. There's a tension "T" in the rope, which pulls down on the cylinder (producing a torque), and pulls up on the hanging mass.

2. Because the hanging mass and the cylinder are connected, there's a relationship between the mass's linear acceleration "a" and the cylinder's angular acceleration "" (alpha). That relationship is:

= a/r (r = cylinder's radius)

Now all that remains is to write "(net force)=ma" for the mass, and "(net torque)=I" for the cylinder.

For the hanging weight ("m" = its mass, 0.770kg)
(net force) = ma
mg - T = ma

For the cylinder:
(net torque) = I
(T)(r) = I

Substitute for :

(T)(r) = Ia/r

For a solid cylinder, I = ½Mr² ("M" = cylinder's mass, 2.07 kg).

So:
(T)(r) = ½Mr²(a/r)

Simplify:
T = ½Ma

Recap:
--------
mg - T = ma
T = ½Ma

You are already given the values of "m" and "M", so these are two equations in two unknowns ("a" and "T"). Use algebra to solve for "a"; then use =a/r to get the angular acceleration.

> "How far does m travel downward..."

Use the formula that relates acceleration, time and distance:

d = ½at² (when starting from rest.

Plug in "t=0.53s" to get the total distance traveled in that time; then plug in "t=0.73s" to get the total distance traveled in THAT time. Then subtract the two distances.

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