In the figure below, one block has mass M = 500.00 g, the other has mass m = 460
ID: 250732 • Letter: I
Question
In the figure below, one block has mass M = 500.00 g, the other has mass m = 460.00 g, and the pulley, which is mounted in horizontal frictionless bearings, has a radius of 5.0000 cm. When released from rest, the more massive block falls 85.800 cm in 2.4700 s (without the cord slipping on the pulley, do not treat the pulley as a uniform disk). (Give your answers to five significant figures. Use these rounded values in subsequent calculations.)
(a) What is the magnitude of the block's acceleration?
m/s2
.281269 (was wrong)
(b) What is the tension in the part of the cord that supports the heavier block?
N
5.04 (was wrong)
(c) What is the tension in the part of the cord that supports the lighter block?
N
4.378 (was wrong)
(d) What is the magnitude of the pulley's angular acceleration?
rad/s2
5.6253 (was right)
(e) What is its rotational inertia? (Give this answer to three significant figures.)
kg · m2
0.00(was wrong)
Explanation / Answer
(a) s = 1/2 at2
Given data : s = 85.8 cm
t = 2.4700 s
so a = (2*s / t^2) = (2*85.8 / (2.47)^2) = 28.127 cm/s^2 = 0.281m/s2
(b) Let m1 = 500g = 0.5 kg
T1 = tension of the string for m1
Then m1g - T1 = m1a
So T1 = m1(g - a)
= (0.5kg) ( (9.8m/s^2) - (0.281m/s^2))
T1 = 4.619 N
(c) Let m2 = 460g = 0.460 kg
T2 is the tension in the string for m2
Then T2 - m2g = m2a ..... the acceleration is the same for both masses
T2 = m2(g+a)
= ( 0.460kg)((9.8m/s^2) + (0.281m/s^2))
T2 = 4.227 N
(d) The angular acceleration ?
? = ?t...(i)
where ? is the angular speed and the initial angular speed is zero
Also, v = ?r = at ===> ? = at/r (ii)
Combining (i) and (ii), we get
?t = at/r
given radius r = 5cm = 0.05m
? = a/r = (0.281 m/s^2) / (0.05 m)
? = 5.62 rad/s2
(d) Rotational inertia I
We use the torque to determine I
? = I?
? = (T1 - T2)r So
I? = (T1 - T2)r
I = (T1 - T2)r/?
I = (4.76N - 4.636N) * (0.05 m) / (5.566 rad/s2)
I = 0.0011139 = 1.114 * 10^(-3) kg.m