A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door

Question

A solid wood door 1.00 m wide and 2.00 m high is hinged along one side and has a total mass of 45.0kg . Initially open and at rest, the door is struck at its center by a handful of sticky mud with mass 0.700 kg, traveling perpendicular to the door at 12.0m/s just before impact
A) Find the final angular speed of the door.
answer in rad/s
B) Does the mud make a significant contribution to the moment of inertia?
Yes or No

in progress 0
King 5 years 2021-08-31T21:03:33+00:00 1 Answers 404 views 0

Answers ( )

  1. Nho
    -4
    2021-08-31T21:05:18+00:00
    Reply

    Answer:

    0.19rad/s and Yes

    Explanation:

    From the principle of conservation of momentum it means momentum before and after collision is the same.

    Momentum before collision is 0.700 kg×12 = 8.4Ns

    Momentum of the door = mass of door × velocity of door

    8.4Ns = mass of door × velocity of door

    Velocity of door = 8.4Ns/45 =0.19m/s

    But velocity V= w×r ;

    w-angular velocity

    r- raduis = width

    w= 0.19/1m = 0.19rad/s

    2. Yes it did because it resisted The moment of inertia and ensued the locking of the door.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )