A turntable must spin at 33.3 rpm (3.49 rad/s) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.50 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.9 cm and mass 0.267 kg.
Question
A turntable must spin at 33.3 rpm (3.49 rad/s) to play an old-fashioned vinyl record. How much torque must the motor deliver if the turntable is to reach its final angular speed in 2.50 revolutions, starting from rest? The turntable is a uniform disk of diameter 30.9 cm and mass 0.267 kg.
Solution
To solve this problem, we need to use the equations of rotational motion.
Step 1: Calculate the moment of inertia (I) of the turntable. The moment of inertia of a disk is given by the formula I = 0.5mr^2, where m is the mass and r is the radius.
Given that the diameter of the turntable is 30.9 cm, the radius r is 30.9/2 = 15.45 cm = 0.1545 m.
So, I = 0.5 * 0.267 kg * (0.1545 m)^2 = 0.00318 kg*m^2.
Step 2: Calculate the angular acceleration (α). The turntable starts from rest and makes 2.5 revolutions to reach its final angular speed.
We know that the final angular speed ω = 3.49 rad/s and the initial angular speed ω0 = 0 rad/s.
We also know that the angular displacement θ = 2.5 revolutions = 2.5 * 2π rad = 15.7 rad.
We can use the equation of motion ω^2 = ω0^2 + 2αθ to solve for α:
(3.49 rad/s)^2 = (0 rad/s)^2 + 2α * 15.7 rad
α = (3.49 rad/s)^2 / (2 * 15.7 rad) = 0.387 rad/s^2.
Step 3: Calculate the torque (τ). The torque is given by the formula τ = Iα.
So, τ = 0.00318 kgm^2 * 0.387 rad/s^2 = 0.00123 Nm.
Therefore, the motor must deliver a torque of 0.00123 N*m for the turntable to reach its final angular speed in 2.50 revolutions, starting from rest.
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