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April 25, 2014

April 25, 2014

Posted by **Kristen** on Sunday, January 27, 2008 at 9:34am.

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?

- AP physics -
**bobpursley**, Sunday, January 27, 2008 at 10:42amYou have distance, time, intial velocity, compute acceleration.

d=1/2 at^2

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.

- AP physics -
**drwls**, Sunday, January 27, 2008 at 10:56amBoth 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.

- AP physics -
**Kristen**, Sunday, January 27, 2008 at 12:25pmI get 4.77 N for T2..do i have a decimal wrong?

- AP physics -
**Jonathan Mad**, Sunday, November 16, 2008 at 5:50pmYeah, 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

- AP physics -
**Jonathan Mad**, Sunday, November 16, 2008 at 5:50pmYeah, 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

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