Question

Four disks of mass m= 50 g radius r= 1.25 cm arranged into an equilateral triangle of side lengths= 3 cm. (One disk is at each corner of the triangle and one is in the center.)

How much torque would you need to apply to flip the spinner over in 0.1 s?

Answer 1

Displacement of one spin, = 2 rad
Consider initial angular velocity, i = 0
Consider the angular acceleration as
Using the relation, = i t + 1/2 t²
= 2/t²
= 4/(0.1)²
= 1.257 x 10³ rad/s².

Moment of inertia of a disk about its center = 1/2 mr²
Using parallel axis theorem, moment of inertia about the center of the equilateral triangle,
I = 1/2 mr² + md²
Where d is the distance from the center of the equilateral triangle to the corner.
d = a/, where a is the side of the equilateral triangle.
d² = a²/3
I = 1/2 mr² + 1/3 ma²
substituting,
I = 50 x 10^-3 x [0.5 x (1.25 x 10^-2)² + 1/3 x (3 x 10^-2)²]
= 1.891 x 10^-5 kg.m²

Moment of inertia of all the three disks, I_net = 3I
= 5.672 x 10^-5 kg.m²

Torque, = I_net
= (1.891 x 10^-5) x (1.257 x 10³)
= 7.13 x 10^-2 Nm