posted by Kristen .
On Block has a mass M=500 g, the other has mass m=460 g, they are hooked to a string and is on a pulley will one mass on either side, the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5 cm. When released from rest, the heavier block falls 81.5 cam in 2.51 s (withouth the cord slipping on the pulley).
What is the magnitude of the blocks acceleration?
What is the tension in the part of the cord that supports the heavier block?
What is the tension in the part of the cord that supports the lighter block?
What is the magnitude of the pulleys angular acceleration in rad/s squared?
What is its rotational inertia?
You have distance, time, intial velocity, compute acceleration.
2j) Tension= mg-acceleration*m
3) Tension= mg+ma
in the case of each of those, m is the mass of the block being investigated.
one knows tangential acceleration, a. angular acceleartion= tangential acceleartion /radius
Finally, w= I*alpha, solve for I.
Both blocks accelerate at the same rate; the lighter one goes up and one the more massive one down. The distance fallen Y in time t obeys the rule
Y = (1/2) a t^2. Therefore a = 2Y/t^2 = 25.9 cm/s^2. (If the pulley had no mass, it would accelerate somewhat faster-- at a rate (M1-M2)g/(M1+M2)= 28.3 cm/s^2, but they don't ask for that). M2 is the mass of the heavier block: 0.5 kg. M1 = 0.46 kg
For the cord tension T2 on the heavier block, solve
M2 g -T2 = M2 a
T2 = M2 (g-a) , where g = 980 cm/s^2 = 9.8 m/s^2. I get 47.7 N
For the tension on M1, solve
M1 g -T1 = -M1 a, since it accelerates upward, in the opposite direction from the gravity force. In this case, T1 = M1 (g+a)
The pulley's angular acceleration is:
alpha = a/R, where R is the pulley's radius.
The rotational moment of inertia (I) of the pulley is given by
Torque = (T1 - T2)*R = I*alpha.
Use the alpha value from the previous step.
I get 4.77 N for T2..do i have a decimal wrong?
Yeah, I got 4.8N. You forgot to convert acceleration (28.3 cm/s^2) into m/s^2. So, (0.5kg)(9.8m/s^2-0.283m/s^2) = T2