Ask Question
28 January, 17:10

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?

+3
Answers (1)
  1. 28 January, 17:27
    0
    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.
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers