A string is wrapped around a uniform disk of mass
M = 1.5 kg
and radius
R = 0.11 m
(see figure below). Attached to the disk are four low-mass rods of radius
b = 0.17 m,
each with a small mass
m = 0.3 kg
at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force
F = 16 N
for a time of 0.2 s. Now what is the angular speed of the apparatus?
rad/s
Also calculate numerically the angle through which the apparatus turns, in radians and degrees.
M, R F2
Solution
torque T = I*alpha
I = (MR^2/2+4mb^2)
I = (1/2*1.5*0.11^2+4*0.3*0.17^2)
I = 0.0438 kg m^2
alpha = 16*0.11/0.0438 = 40.18 rad/s^2
at t = 0.2 sec
angular velocity w2 = w1+alpha*t
w2 = 0+40.18*0.2 = 8.036 rad/s
from the relation
theta = w1*t+1/2*alpha*t^2
theta = 1/2*40.18*0.2^2 = 0.8036 rad
theta = 46.04 degrees
.
A string is wrapped around a uniform disk of mass M - 1-5 kg and radiu.docx
1. A string is wrapped around a uniform disk of mass
M = 1.5 kg
and radius
R = 0.11 m
(see figure below). Attached to the disk are four low-mass rods of radius
b = 0.17 m,
each with a small mass
m = 0.3 kg
at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string
with a constant force
F = 16 N
for a time of 0.2 s. Now what is the angular speed of the apparatus?
rad/s
Also calculate numerically the angle through which the apparatus turns, in radians and degrees.
? ? radians = rad
? ? degrees = °
M, R F2
Solution
torque T = I*alpha
2. I = (MR^2/2+4mb^2)
I = (1/2*1.5*0.11^2+4*0.3*0.17^2)
I = 0.0438 kg m^2
alpha = 16*0.11/0.0438 = 40.18 rad/s^2
at t = 0.2 sec
angular velocity w2 = w1+alpha*t
w2 = 0+40.18*0.2 = 8.036 rad/s
from the relation
theta = w1*t+1/2*alpha*t^2
theta = 1/2*40.18*0.2^2 = 0.8036 rad
theta = 46.04 degrees
