In the figure shown one block has mass m 1 = 500 g, the other has a mass m 2 = 4
ID: 1397250 • Letter: I
Question
In the figure shown one block has mass m
1 = 500 g, the other has a mass m2 = 460 g, and the pulley which is mounted in horizontal frictionless bearing, has a radius of 5.00 cm. When released from rest, mass m1 falls 75.0 cm in 5.00 s.
What is the magnitude of the block’s acceleration?
What is the magnitude of tension T1?
What is the magnitude of tension T2?
What is the magnitude of the pulley’s angular acceleration?
What is the magnitude of the net torque acting on the pulley?
What is the pulley’s moment of inertia?
Explanation / Answer
alon vertical
displacement y = -0.75 m
time t = 5 s
initial velocity voy = 0
from equations of motion
y = voy*t + 0.5*ay*t^2
-0.75 = 0 - 0.5*ay*5^2
ay = 0.06 m/s^2 ,----answer
part(b)
m1 moving down wards
a = -0.06 m/s^2
Fnet = T1 - m1*g
from newtons law Fnet = m*a
T1 - m1*g = m1*a
t1 = m1 *(g+a)
T1 = 0.5*(9.8-0.06) = 4.87 N <-----answer
-----------
part(c)
m2 is moving up
a = + 0.06 m/s^2
Fnet = T2 - m2*g
from newtons law
Fnet = m2*a
T2 = m2*a + m2*g
T2 = m2*(g+a)
T2 = 0.46*(9.8+0.06)
T2 = 4.5356 6N <----answer
---------------
angular acceleration = alfa = a/r = 0.06/0.05 = 1.2 rad/s^2 <---------answer
-----------
net torque = (T1 - T2)*r
net torque = (4.87-4.5356)*0.05 = 0.01672 Nm <---------answer
====================
for pulley
net torque = I*alfa
(T1 - T2)*r = I*alfa
(4.87-4.5356)*0.05 = I*1.2
I = 0.014 kg m^2 <---------answer
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