Question

A 460 g, 8.5-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.4 m/s.

Answer 1

To find the moment of inertia of the can, we need to know its shape and axis of rotation. Assuming the can is a solid cylinder and rolls along its longitudinal axis, the moment of inertia can be calculated as:

I = 1/2 * m * r^2

where I is the moment of inertia, m is the mass of the can, and r is the radius of the can.

Given:
m = 460 g = 0.46 kg
r = 8.5 cm = 0.085 m

Plugging these values into the formula, we get:

I = 1/2 * 0.46 kg * (0.085 m)^2
= 0.001589 kg*m^2

Now, we can use the concept of rotational kinetic energy to find the can's rotational speed. The rotational kinetic energy (K_rot) is given by:

K_rot = 1/2 * I * ω^2

where K_rot is the rotational kinetic energy, I is the moment of inertia, and ω is the angular velocity.

Given:
K_rot = K_trans = 1/2 * m * v^2 (due to uniform, dense food)
m = 0.46 kg
v = 1.4 m/s

Plugging these values into the equation, we get:

0.001589 kg*m^2 * ω^2 = 1/2 * 0.46 kg * (1.4 m/s)^2

Simplifying and solving for ω, we find:

ω^2 = (1/2 * 0.46 kg * (1.4 m/s)^2) / 0.001589 kg*m^2

ω^2 = 0.563

Applying square root on both sides of the equation, we get:

ω = √0.563

Taking the square root of 0.563, we find:

ω ≈ 0.751 rad/s

Therefore, the can is rolling with an angular velocity of approximately 0.751 rad/s.