Posted by Brit on Sunday, February 26, 2012 at 4:35pm.
A 3.00 kg stone is tied to a thin, light wire wrapped around the outer edge of the uniform 8.00 kg cylindrical pulley. The inner diameter of the pulley is 60.0 cm, while the outer diameter is 1.00 m. The system is released from rest, and there is no friction at the axle of the pulley. Find (a) the acceleration of the stone, (b) the tension in the wire, and (c) the angular acceleration of the pulley.
for part a i did
I of the pulley is
I = m*(r1^2 + r2^2)/2)
I = 8*(0.5^2 + 0.3^2)/2)
I =1.36 kgm^2
a = 3*9.81/(3 + 1.36/0.5^2)
a = 3.48 m/s^2 <--- acceleration of the stone
á = a/R = 3.48/0.5 = 6.96 rad/s^2 <--- angular acceleration
T = mg - ma = 3(9.81-3.48)=18.99 N <--- tension
- Physics please help! check if its correct - Brit, Sunday, February 26, 2012 at 4:37pm
B and C is switched anwsers
- Oh my - sign ! - Damon, Sunday, February 26, 2012 at 5:42pm
The inner radius I assume means that the axle does not spin. Therefore your moment of inertia has a negative, not positive sign
I = 8*(0.5^2 - 0.3^2)/2)
I =.64 kgm^2
- Physics please help! check if its correct - Damon, Sunday, February 26, 2012 at 5:48pm
stone F = m a
a = (mg-T)/ m = (3*9.81 -T)/3
wheel Torque = I alpha
alpha = .5 T/.64
a = alpha * r
a = .25 T/.64
(3*9.81 -T)/3 = .25 T/.64
solve for T
go back and solve for a
and for alpha = a/r
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