An irregularly shaped plastic plate with uniform thickness and density (mass per unit volume) is to be rotated around an axle that is perpendicular to the plate face and through point O. The rotational inertia of the plate about that axle is measured with the following method. A circular disk of mass 0.500 kg and radius 2.00 cm is glued to the plate, with its center aligned with point O. A string is wrapped around the edge of the disk the way a string is wrapped around a top. Then the string is pulled for 5.00 s. As a result, the disk and plate are rotated by a constant force of 0.400 N that is applied by the string tangentially to the edge of the disk. The resulting angular speed is 114 rad/s. What is the rotational inertia of the plate about the axle?
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Since this is an irregularly shaped object which equation for inertia would I use? I = 1/2*M*R^2???
Would I start out with T = r*F*sin theta, with theta = 90 so sin theta = 1.
T = r*F, so T = (0.02)(0.400) = 0.008. Then what do I do?
To determine the rotational inertia of the plate about the axle, we can use the information given about the circular disk that is glued to the plate. Since the disk's mass and radius are given, we can calculate its rotational inertia separately and then subtract it from the total rotational inertia of the plate and disk system.
The rotational inertia of a circular disk about an axis perpendicular to its face and passing through its center is given by the formula:
I_disk = (1/2) * m * r^2
where m is the mass of the disk and r is the radius of the disk.
In this case, the mass of the disk is 0.500 kg and the radius is 2.00 cm (or 0.02 m). Substituting these values into the formula, we have:
I_disk = (1/2) * 0.500 kg * (0.02 m)^2
= 0.005 kg * m^2
Therefore, the rotational inertia of the disk is 0.005 kg * m^2.
To find the rotational inertia of the plate about the axle, we subtract the rotational inertia of the disk from the total rotational inertia of the plate and disk system. The total rotational inertia is given by:
I_total = I_plate + I_disk
We are given the angular speed of the plate and disk system, which is 114 rad/s. We can relate the applied torque (T) to the rotational inertia (I) and the angular acceleration (α) using the equation:
T = I * α
Since the force applied tangentially to the edge of the disk provides the torque, we can calculate it using:
T = r * F
where r is the radius of the disk.
Given that the radius of the disk is 0.02 m and the applied force is 0.400 N, we have:
T = (0.02 m) * (0.400 N)
= 0.008 N*m
Now, we need to find the angular acceleration (α) to use in the equation. We can use the relationship between angular acceleration, angular velocity, and time:
α = Δω / Δt
where Δω is the change in angular velocity and Δt is the time interval.
We are told that the angular speed increases from 0 rad/s to 114 rad/s in 5.00 s. Thus, the change in angular velocity is:
Δω = 114 rad/s - 0 rad/s
= 114 rad/s
Substituting this value and the time interval into the equation, we have:
α = (114 rad/s) / (5.00 s)
= 22.8 rad/s^2
Now, we can rearrange the equation T = I * α to solve for the rotational inertia of the plate:
I_plate = T / α
Substituting the values we have calculated, we get:
I_plate = (0.008 N*m) / (22.8 rad/s^2)
≈ 0.00035 kg * m^2
Therefore, the rotational inertia of the plate about the axle is approximately 0.00035 kg * m^2.