Two wheels have the same mass and radius of 4.0 kg and 0.35 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 13 rad in 8.0 s. Find the net external torque that acts on each wheel.
They start standing still and accelerate to turn 13.0 rad in 8.0 seconds.
The average angular velocity during that interval is 13/8 = 1.625 rad/s. The maximum velocity achieved is twice that, 3.25 rad/s. The angular acceleraqtion rate is 3.25/8 = 0.4063 rad/s^2.
The torque is the angular acceleration rate divided by the moment of inertia, I. For the hoop, I = M R^2. For the disc, I = (1/2)MR^2.
The two wheels have different moments of inertia.
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An inverted “V” is made of uniform boards and weighs 373 N. Each side has the same length and makes a 30.0° angle with the vertical, as the figure shows. Find the magnitude of the static frictional force that acts on the lower end of each leg of the “V.”
To find the net external torque that acts on each wheel, we can use the equation:
τ = Iα,
where τ represents the net external torque, I is the moment of inertia of the wheel, and α is the angular acceleration.
For wheel (a) with the shape of a hoop, the moment of inertia can be calculated using the formula:
I = mR^2,
where m is the mass of the wheel and R is the radius. Plugging in the values, we get:
I(a) = 4.0 kg * (0.35 m)^2 = 0.49 kg.m^2.
For wheel (b) with the shape of a solid disk, the moment of inertia can be calculated using the formula:
I = (1/2) mR^2,
where m is the mass of the wheel and R is the radius. Plugging in the values, we get:
I(b) = (1/2) * 4.0 kg * (0.35 m)^2 = 0.343 kg.m^2.
Next, we need to find the angular acceleration α. We know that each wheel turns through an angle of 13 rad in 8.0 s. The relationship between angular acceleration, angle, and time is given by:
θ = (1/2) α t^2,
where θ is the angle, α is the angular acceleration, and t is the time. Rearranging the equation, we have:
α = (2θ) / t^2.
Plugging in the values, we get:
α = (2 * 13 rad) / (8.0 s)^2 = 0.405 rad/s^2.
Now, we can calculate the net external torque for each wheel using the equation τ = Iα.
For wheel (a):
τ(a) = I(a) * α = 0.49 kg.m^2 * 0.405 rad/s^2 = 0.19845 N.m.
For wheel (b):
τ(b) = I(b) * α = 0.343 kg.m^2 * 0.405 rad/s^2 = 0.139215 N.m.
Therefore, the net external torque that acts on wheel (a) is 0.19845 N.m and the net external torque that acts on wheel (b) is 0.139215 N.m.