I just need some clarification on part (c). Here's the whole problem for context:
A uniform solid cylinder of mass 10 kg can rotate about a frictionless axle through its center O, as shown in the cross-sectional view in the figure. A rope wrapped around the outer radius R1 = 1.0 m exerts a force of magnitude F1 = 5.0 N to the right. A second rope wrapped around another section of radius R2 = 0.50 m exerts a force of magnitude F2 = 6.0 N downward. (The moment of inertia of a uniform solid cylinder is I = ½ m r^2)
a) What is the angular acceleration of the cylinder?
T = (F1*R1)-(F2*R2) = (1/2)mr^2(alpha)
(5*1)-(6*.50)-/(.5)(10)(1) = 0.4
angular acceleration = 0.4 rad/s^2
b) How many revolutions does the cylinder rotate through the first 5.0 seconds if it starts from rest?
theta = w_0 + .5(angular acceleration)(t)^2
0 + .5(.4)(5)^2 = 5.0 rad
c) Which way does it rotate, clockwise or counterclockwise?
I want to say counterclockwise, but the key I have says it should be clockwise. I thought it should be counterclockwise because the displacement is positive. But I'm not sure if that reasoning is right. Can you explain why it's clockwise? It's the only part of this problem I don't understand.
(cw = clockwise, ccw = counterclockwise). It should be clockwise since it is negative rotation. The force has Torque(1) = -5 meaning it rotates cw and Torque(2) = 3 meaning it rotates ccw. T(1) + T(2) = -2 so overall it is clockwise.
To determine the direction of rotation, you can use the right-hand rule for rotation. Here's how you can apply it to this problem:
1. Imagine holding the cylinder with your right hand such that your fingers are wrapped around the axle, and your thumb points in the direction of the positive angular velocity.
2. Now, consider the forces acting on the cylinder:
- The force F1 to the right tries to rotate the cylinder counterclockwise (in the direction of your fingers).
- The force F2 downward also tries to rotate the cylinder counterclockwise (in the direction of your fingers).
3. Since both forces are acting in the counterclockwise direction, their torques on the cylinder add up, causing a net counterclockwise torque. According to Newton's second law for rotational motion, this net torque will result in a clockwise angular acceleration.
Therefore, the cylinder rotates clockwise.
To determine the direction of rotation, you can use the concept of torque. Torque is the rotational equivalent of force and determines the tendency of an object to rotate. In this case, the clockwise or counterclockwise rotation depends on the net torque acting on the cylinder.
Let's consider the torques due to each force. The force F1 exerts a torque in the counterclockwise direction, as it is applied horizontally to the right of the axis of rotation. The torque due to F1 is given by τ1 = F1 * R1.
On the other hand, the force F2 exerts a torque in the clockwise direction, as it is applied vertically downward, which is opposite to the counterclockwise direction. The torque due to F2 is given by τ2 = -F2 * R2. The negative sign indicates the opposite direction.
To determine the net torque, you need to sum up these torques. Since τnet = τ1 + τ2, substituting the values: τnet = (F1 * R1) + (-F2 * R2).
Now, let's substitute the given values:
τnet = (5.0 N * 1.0 m) + (-(6.0 N) * 0.50 m)
= 5.0 Nm - 3.0 Nm
= 2.0 Nm
The net torque is positive (2.0 Nm), indicating a tendency for clockwise rotation. Therefore, the cylinder will rotate in the clockwise direction.
Remember that torque is analogous to force, and its direction is determined using the right-hand rule. When applying the right-hand rule, the thumb points in the direction of the positive torque, and the curled fingers indicate the rotation direction. In this case, the direction of the torque is clockwise, which means the cylinder will rotate in that direction.
Hence, the key is correct in stating that the cylinder will rotate clockwise.