Question

Part 1. A uniform hoop rolls without slipping down a 10° inclined plane. What is the acceleration of the hoop's center of mass?

The moment of inertia of a uniform solid disk about an axis that passes through its center = mr².
The moment of inertia of a uniform solid disk about an axis that is tangent to its surface = 2mr².

Part 2. A solid disk rotates in the horizontal plane at an angular velocity of 2.50×10−22.50×10−2 rad/s with respect to an axis perpendicular to the disk at its center. The moment of inertia of the disk is 0.12 kg.m2. From above, sand is dropped straight down onto this rotating disk, so that a thin uniform ring of sand is formed at a distance of 0.40 m from the axis. The sand in the ring has a mass of 0.50 kg. After all the sand is in place, what is the angular velocity of the disk? (in rad/s)

Answer 1

1.
`a = 0.87m/s^2`

2.
`\omega_2= 1.5*10^(-2)rad/s`

Step-by-step explanation:

For part 1, we need to make a sum of torque at the rotating point:

`m*g*sin(\theta)*r=I*\alpha=2*m*r^2*(a/r)`

Solving for a:

`a = g*sin(\theta)/2 = 0.87m/s^2`

For part 2, we need the new moment of inertia with the sand ring:

`I1 = 0.12 kg.m^2`

`I2=I1+m2*r2^2=0.2 kg.m^2`

By conservation of the angular momentum:

`I1*\omega_1=I2*\omega_2`

`\omega_2=I1*\omega_1/I2=1.5*10^(-2)rad/s`