Multiple-Concept Example 10 provides one model for solving this type of problem. Two wheels have the same mass and radius of 5.0 kg and 0.47 m, respectively. One has (a) the shape of a hoop and the other (b) the shape of a solid disk. The wheels start from rest and have a constant angular acceleration with respect to a rotational axis that is perpendicular to the plane of the wheel at its center. Each turns through an angle of 12 rad in 9.3 s. Find the net external torque that acts on each wheel.
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To find the net external torque that acts on each wheel, we can use the rotational kinematic equation:
θ = ω₀t + (1/2)αt²
where θ is the angle turned, ω₀ is the initial angular velocity, α is the angular acceleration, and t is the time taken.
Given that both wheels start from rest and have a constant angular acceleration, we can write the equation for both wheels as:
θ = (1/2)αt²
Since the angle turned (θ) and the time taken (t) are the same for both wheels, we can equate the equations for the hoop (a) and the solid disk (b) to find the ratio of their angular accelerations (α):
(1/2)α(a)t² = (1/2)α(b)t²
The mass and radius of the wheels are the same, so the only difference between the hoop and the solid disk is their moment of inertia (I).
The moment of inertia for a hoop is I(a) = m(a)r², where m(a) is the mass of the hoop and r is the radius.
The moment of inertia for a solid disk is I(b) = (1/2)m(b)r², where m(b) is the mass of the disk and r is the radius.
Since the masses are the same, we can write:
I(a) = I(b)
m(a)r² = (1/2)m(b)r²
Now we can solve for the ratio of their angular accelerations (α):
(1/2)α(a)t² = (1/2)α(b)t²
α(a) = α(b)
So, the angular accelerations (α) for both wheels are the same.
Since the net external torque (τ) is related to the moment of inertia (I) and angular acceleration (α) by the equation:
τ = Iα
we can conclude that the net external torque acting on both the hoop and the solid disk is the same.