A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed

Question

A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal disk is placed in contact with a wall, and the disk comes to rest after 10s. Which of the following situations associated with linear impulse is analogous to the angular impulse that is described?

a. A 3kg block is initially at rest. An applied force of 3N is applied to the block, but the block does not move.
b. A 3kg block is initially at rest. A net force of 3N is applied to the block until it has a speed of 10m/s.
c. A 3kg block is initially traveling at 10m/s. An applied force of 3N is applied to the block in the direction of its velocity vector for 10s.
d. A 3kg block is initially traveling at 10m/s. The block encounters a 3N frictional force until the block eventually stops.

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Huyền Thanh 5 years 2021-09-01T01:54:57+00:00 1 Answers 27 views 0

Answers ( )

  1. thachthao
    0
    2021-09-01T01:56:42+00:00
    Reply

    Answer:

    D

    Explanation:

    From the information given:

    The angular speed for the block \omega = 50 \ rad/s

    Disk radius (r) = 0.2 m

    The block Initial velocity is:

    v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s

    Change in the block’s angular speed is:

    \Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

    However, on the disk, moment of inertIa is:

    I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

    The time t = 10s

    ∴

    Frictional torques by the wall on the disk is:

    T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m

    Finally, the frictional force is calculated as:

    F = \dfrac{T}r{}

    F= \dfrac{0.6}{0.2} \\ \\ F = 3N

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