Suppose you’re eating in a restaurant where the dishes are shared at the table and all placed uniformly on a rotating disk-like surface. Model this surface as a thin disk of radius 33.4 cm. You can’t stop thinking about physics even though you’re out with your friends, and decide to calculate the mass of the rotating surface and all the food. If the surface is initially at rest and you exert a tangential force of 1.7 N on it, you observe that the food rotates at a speed of 1 rev/s after applying the force consistenty for 1.3 seconds. Find the mass of the disk wtih the food, in kg.

Answer 1

First, let's find the rotational inertia (moment of inertia) of the disk with the food.

The moment of inertia of a thin disk is given by the formula: I = (1/2) * M * R^2,
where I is the moment of inertia, M is the mass, and R is the radius of the disk.

Since we're given the angular speed, let's first convert the 1 rev/s to rad/s.
1 rev = 2π rad
So, 1 rev/s = 2π rad/s

Now we can use the rotational kinematic equation to find the angular acceleration:
ω^2 = ω_initial^2 + 2 * α * θ

The disk starts from rest, so ω_initial = 0. The angle (θ) covered by the disk during 1.3 seconds is:
θ = ω * t = 2π (1.3) = 2.6π rad

Now we can plug these into the equation and solve for angular acceleration (α):
(2π)^2 = 0 + 2 * α * 2.6π
4π^2 = 5.2π * α
α = 4π^2 / 5.2π = 8π / 2.6 = 4π / 1.3 rad/s^2

Now, let's use the torque equation to find the moment of inertia I:
τ = I * α

The torque (τ) applied by the tangential force is: τ = F * R = 1.7 N * 0.334 m = 0.568 N.m
Thus, 0.568 N.m = I * (4π / 1.3)
I = 0.568 * 1.3 / 4π = 0.1465 kg.m^2

Now that we have the moment of inertia (I), we can find the mass of the disk with the food using the formula I = (1/2) * M * R^2:
0.1465 kg.m^2 = (1/2) * M * (0.334 m)^2
M = 2 * 0.1465 / (0.334^2) = 2.601 kg

So, the mass of the disk with the food is approximately 2.6 kg.