I got the first three right, but put them in here to make it easier. The fourth

Question

I got the first three right, but put them in here to make it easier. The fourth part is what I cannot figure out!

M, a solid cylinder (M=1.39 kg, R=0.111 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. Calculate the angular acceleration of the cylinder.

9.79×101 rad/s^2

If instead of the force F an actual mass m = 0.770 kg is hung from the string, find the angular acceleration of the cylinder.

4.65×101 rad/s^2

How far does m travel downward between 0.450 s and 0.650 s after the motion begins?
5.67×10-1 m

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 0.677 m in a time of 0.570 s. Find Icm of the new cylinder.

.0223 kg m^2 <-- this was the answer I put and it is incorrect.

This is the "hint" it gave me --> You know how to write an equation for the torque from the second part. You are given quantities to calculate the acceleration (review motion in one dimension if this is not obvious), and in this case you need to solve for the moment of inertia.

Explanation / Answer

ANSWER

Part A

T*r = I*

so, T*r = 0.5*m*r^2 *

so, = 2T/mr

also T = F=7.554N

so = 97.92 rad/s2   ANS.



Part B

If an actual mass is hung then by drawing the free body diagram of the mass, we get:

mg - T = ma

also T*r = I* = 0.5*M*r^2 *a/r

so, T =0.5Ma

so, substituting the force equation,we get

mg - Ma/2 = ma

so , (m+M/2)a = mg

so, a= mg/(m+M/2) =5.151 m/s2

so, angular acceleration = a/r = 46.40 rad/s2   ANS.



Part C

a =5.151 m/s2 (obtained from above)

using the equation, s =u*t +0.5*a*t^2

for start from rest, u =0

so, s= 0.5*a*t^2

so, s1= 0.5*a*t1^2

and , s2 = 0.5*a*t2^2

so, s2-s1 = 0.5*a*(t2^2-t1^2) = 0.567m   ANS.

Part D

Using the same equation

0.677 = 0.5*a*0.570^2
so, a = 4.167 m/s2



solving backwards, we get
T*r = I*a/r
so, T = I*a/r^2

substituting in the main force equation, we get,
m*g - I*a/r^2 = m*a
so, I*a/r^2 = m(g-a)

so, I = m(g-a)*r^2/a = 0.0128 kg*m2   ANS.

Regards!!!

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