A one-piece cylinder is shaped as in the Figure, with a core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing. A rope wrapped around the drum, of radius 2.70 m, exerts a force F1 to the right on the cylinder. A rope wrapped around the core, of radius 1.67 m, exerts a force F2 downward on the cylinder. Let F1 = 5.00 N, F2 = 6.70 N and the moment of inertia of the system I = 0.132 kg m2.
a) What is in N m the net torque acting on the cylinder about the rotation axis (which is the z axis in the Figure)? Note: CCW rotation is considered positive, i.e. the torque vector points in the positive z direction.
I can't figure out how to include the inertia into the torque equation T= F2r2 -f1r1. F2 is pointing in the counterclockwise direction and f2 is pointing in the clockwise direction.
b) What would be the magnitude of the rotational velocity of the cylinders at t = 9.30 s? Express your answer in rad/s
To determine the net torque acting on the cylinder, we need to consider both the torque exerted by the force F1 and the torque exerted by the force F2.
The equation for torque is given by:
Torque = Force x Distance x sin(θ)
For F1, the force is 5.00 N and the distance from the axis of rotation to the point where the force is applied is the radius of the drum, which is 2.70 m. As the force F1 is exerted to the right, the angle θ is 0 degrees (or π radians).
For F2, the force is 6.70 N and the distance from the axis of rotation to the point where the force is applied is the radius of the core, which is 1.67 m. As the force F2 is exerted downward, the angle θ is 90 degrees (or π/2 radians).
Now we can calculate the torques:
Torque1 = F1 x r1 x sin(θ1) = 5.00 N x 2.70 m x sin(0) = 0 Nm (as sin(0) = 0)
Torque2 = F2 x r2 x sin(θ2) = 6.70 N x 1.67 m x sin(π/2) = 6.70 Nm
To find the net torque, we need to find the vector sum of these torques since they act in different directions. The torque vector from F1 is zero, as it is in the positive z direction, and the torque vector from F2 is 6.70 Nm in the negative z direction.
Therefore, the net torque acting on the cylinder about the z axis is -6.70 Nm.
For part (b), to find the magnitude of the rotational velocity of the cylinder at t = 9.30 s, we need to use the equation:
Net torque = Moment of inertia x Angular acceleration
We know the net torque, which is -6.70 Nm, and the moment of inertia, which is 0.132 kg m^2, so we need to find the angular acceleration.
Rearranging the equation, we get:
Angular acceleration = Net torque / Moment of inertia
Angular acceleration = (-6.70 Nm) / (0.132 kg m^2) = -50.76 rad/s^2
To find the rotational velocity at t = 9.30 s, we need to integrate the angular acceleration with respect to time.
ω = ω0 + α * t
where:
ω0 = initial rotational velocity (which is most likely zero in this case)
α = angular acceleration
t = time
Since the initial rotational velocity is assumed to be zero, we have:
ω = 0 + (-50.76 rad/s^2) * 9.30 s
ω = -472.43 rad/s
Since the magnitude is being asked for, we take the absolute value:
Magnitude of rotational velocity = 472.43 rad/s
Therefore, the magnitude of the rotational velocity of the cylinder at t = 9.30 s is 472.43 rad/s.
To find the net torque acting on the cylinder about the rotation axis, we can use the equation:
Net Torque = (F2 * r2) - (F1 * r1)
where F1 and F2 are the forces exerted by the ropes, and r1 and r2 are the radii of the drum and the core respectively.
In this case, F1 = 5.00 N, F2 = 6.70 N, r1 = 2.70 m, and r2 = 1.67 m.
Substituting these values into the equation, we get:
Net Torque = (6.70 N * 1.67 m) - (5.00 N * 2.70 m)
Simplifying, we find:
Net Torque = 11.139 N·m - 13.50 N·m
Net Torque = -2.361 N·m (rounding to 3 decimal places)
So, the net torque acting on the cylinder is -2.361 N·m.
b) To find the magnitude of the rotational velocity of the cylinder at t = 9.30 s, we can use the equation:
Torque = I * angular acceleration
Given the moment of inertia of the system (I) as 0.132 kg·m^2, we need to find the angular acceleration.
The angular acceleration can be found using the equation:
Torque = I * angular acceleration
Rearranging the equation, we get:
Angular acceleration = Torque / I
Substituting the known values, we have:
Angular acceleration = (-2.361 N·m) / (0.132 kg·m^2)
Simplifying, we find:
Angular acceleration = -17.886 rad/s^2 (rounding to 3 decimal places)
Now, to find the magnitude of the rotational velocity at t = 9.30 s, we use the equation:
Angular velocity (ω) = Initial angular velocity (ω₀) + (Angular acceleration * time)
Assuming the initial angular velocity is 0 (since it is not given in the question), we have:
Angular velocity (ω) = (Angular acceleration * time) = -17.886 rad/s^2 * 9.30 s
Calculating, we find:
Angular velocity (ω) = -166.735 rad/s (rounding to 3 decimal places)
Therefore, the magnitude of the rotational velocity of the cylinder at t = 9.30 s is 166.735 rad/s.