You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend pushes the edge of the merry go round with 100 N of force, applied at a 90 deg angle. (hint: the merry-go-round is a disk).

How much torque did your friend apply?

Explanation:

Incomplete question

This is the complete question

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend pushes the edge of the merry go round with 100 N of force, applied at a 90 deg angle. (hint: the merry-go-round is a disk).

a. How much torque did your friend apply?

b. What is your angular acceleration if you are sitting in the middle of the merry-go-round?

c. What is your angular acceleration if you are sitting at the edge of the merry-go-round?

d. Which angular acceleration is larger and why?

Given that,

Mass of Merry go round

M = 200kg

Radius of Merry go round

R = 2m

Mass of person on sitting on the merry go round

m = 50kg

Force applied to push the merry go round

F = 100N

The force is applied at an angle of 90°

θ = 90°

a. Torque?

The torque can calculated using the formula

τ = R × F. This is the cross product of R and F

τ = R•F•Sinθ

Then,

τ = 2 × 100 × Sin90

τ = 200 Nm

b. Angular acceleration if the person is sitting at the middle

Since the boy is sitting at the middle, then is distance from the centre is r' = 0

We know that

τ = I•α

Where, I is moment of inertial and α is angular acceleration

So, we need to find the moment of inertial

Moment of inertial is calculated using

I = MR², since the merry go round is model as a disk

Then,

Total moment of inertial of the boy and the merry go round

I = MR² + mr'²

I = 200 × 2² + 50 × 0²

I = 800 kgm²

So, τ = I•α

α = τ / I

α = 200/800

α = 0.25 rad/s²

c. Angular acceleration if the person is sitting at the edge

Since the person is sitting at the edge, then is distance from the centre is r = R = 2m

We know that

τ = I•α

Where, I is moment of inertial and α is angular acceleration

So, we need to find the moment of inertial

Moment of inertial is calculated using

I = MR², since the merry go round is model as a disk

Then,

Total moment of inertial of the boy and the merry go round

I = MR² + mR²

I = 200 × 2² + 50 × 2²

I = 800 + 200

I = 1000kgm²

So, τ = I•α

α = τ / I

α = 200/1000

α = 0.2 rad/s²

d. The angular acceleration when the person is sitting at the center is greater than when the person is sitting at the edge.

This is due to that the boy does not have any moment of inertial at the center. Also, at the edge the boy has a moment inertial. So, the moment of inertia at the edge is greater than the moment of inertial at the center

Then, we know that, α = τ / I

So, angular acceleration is inversely proportional to the moment of inertial, this implies that at the moment of inertial increases the angular acceleration reduces and as the moment of inertia decrease the angular acceleration decreases.

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.