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.
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.
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
To find the answers to these questions, we can use Newton's laws of motion and the principles of rotational motion.
1. Magnitude of the blocks acceleration:
We can use the kinematic equation for linear motion: d = 1/2 * a * t^2, where d is the distance fallen by the heavier block, a is the acceleration, and t is the time taken. Rearranging the equation, we have: a = (2 * d) / (t^2). Substituting the given values, we have a = (2 * 0.815 m) / (2.51 s^2) = 0.650 m/s^2.
2. Tension in the part of the cord that supports the heavier block:
The tension in the cord is equal to the force exerted on the block. From Newton's second law of motion, F = m * a, where F is the force, m is the mass of the block, and a is the acceleration. Substituting the values, we have F = (0.5 kg) * (0.650 m/s^2) = 0.325 N.
3. Tension in the part of the cord that supports the lighter block:
Since the system is assumed to be in equilibrium, the tension in the cord supporting the lighter block is equal to the weight of the lighter block. Tension = mg = (0.46 kg) * (9.8 m/s^2) = 4.508 N.
4. Magnitude of the pulley's angular acceleration:
Considering the rotational dynamics, we can relate the linear acceleration of the blocks and the angular acceleration of the pulley. The blocks are connected by a string, so as one block falls, the other block is pulled up, creating an angular acceleration in the pulley. The angular acceleration is related to the linear acceleration by the equation: a = r * α, where a is the linear acceleration, r is the radius of the pulley, and α is the angular acceleration. Rearranging the equation, we have: α = a / r = (0.650 m/s^2) / (0.05 m) = 13 rad/s^2.
5. Rotational inertia of the pulley:
The inertia of an object rotating about its axis is called rotational inertia (or moment of inertia). For a disk-shaped pulley, the moment of inertia is given by: I = (1/2) * m * r^2, where m is the mass of the pulley and r is its radius. Substituting the values, we have I = (1/2) * (0.05 kg) * (0.05 m)^2 = 6.25 x 10^-5 kg m^2.
Note: In these calculations, we assumed no slippage and negligible friction.