Consider a 6·kg disk with a radius of 0.6·m and rotating about an axis passing through its center.
(a) What is its rotational inertia? kg·m2
(b) Suppose the disk is rotating counterclockwise and slowing down with a rotational acceleration of 5·rad/s2. What is the net torque acting on the disk? N·m
(c) Suppose the torque you found for the last part is being produced by a force applied at a point 0.3·m from the axis. If the direction of the force is perpendicular to a line from the axis to that point, what is the magnitude of the force? N
Rotational inertia I=mr^2 while m is 6 kg and r is 0.6 m
B) torque =Ia while a is -5 rad/s
To answer these questions, we need to use the formulas related to rotational motion. Let's go step by step:
(a) The rotational inertia of a disk rotating about an axis passing through its center can be found using the formula:
I = (1/2) * m * r^2
where I is the rotational inertia, m is the mass of the disk, and r is the radius of the disk.
In this case, the mass of the disk is given as 6 kg and the radius is given as 0.6 m. Plugging these values into the formula:
I = (1/2) * 6 kg * (0.6 m)^2
I = 1.8 kg·m^2
So, the rotational inertia of the disk is 1.8 kg·m^2.
(b) The net torque acting on the disk can be found using the formula:
τ = I * α
where τ is the net torque, I is the rotational inertia, and α is the rotational acceleration.
In this case, the rotational inertia of the disk is 1.8 kg·m^2 and the rotational acceleration is given as 5 rad/s^2. Plugging these values into the formula:
τ = 1.8 kg·m^2 * 5 rad/s^2
τ = 9 N·m
So, the net torque acting on the disk is 9 N·m.
(c) The magnitude of the force can be found using the formula:
τ = r * F * sin(θ)
where τ is the torque, r is the distance from the axis to the point of application of the force, F is the magnitude of the force, and θ is the angle between the line from the axis to the point and the direction of the force.
In this case, the torque is given as 9 N·m and the distance r is given as 0.3 m. The force is perpendicular to the line from the axis to the point, so sin(θ) becomes 1. Plugging these values into the formula:
9 N·m = 0.3 m * F * 1
F = 9 N·m / 0.3 m
F = 30 N
So, the magnitude of the force is 30 N.