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

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maxwellekoh maxwellekoh · Answer 1 · 2020-06-10T21:25:42+02:00

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.