Ask a New Question

Question

A 346 kg merry-go-round in the shape of a
horizontal disk with a radius of 1.7 m is set in
motion by wrapping a rope about the rim of
the disk and pulling on the rope.
How large a torque would have to be exerted to bring the merry-go-round from rest
to an angular speed of 4.7 rad/s in 3.4 s?
12 years ago

Answers

Damon
angular acceleration = a = 4.7/3.4 = 1.38 radians/s^2

I = (1/2)mr^2 = (1/2)346(1.7^2) = 500kg m^2

Torque = I a = 500 (1.38) = 690 Nm
12 years ago

Related Questions

A 175 kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in... A merry-go-round accelerates from est at .870 rad/sec^2. When the merry-go-round has made 4 complete... A merry go round has a radius of 3.5 meters. How far does a child standing at the edge travel (path... A 4.4 diameter merry-go-round is rotating freely with an angular velocity of 0.63 . Its total momen... A merry-go-round with r = 4m and a perfect frictionless bearing is pushed with a force of 24 N by a... A playground merry-go-round has a radius of R= 2 m and has a moment of inertia Icm= 9×103kg⋅m2 abou... A playground merry-go-round of radius R = 1.1 m has a moment of inertia of I = 280 kg*m^2. and is ro... A merry go round take 5 second to complete 1 revolution find A, The angular velocity in (rad/sec)... Jane is on a merry go round that is moving and it comes with speed. Is her velocity also consistent...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use