An amusement park ride has a radius of 4.2 m. What angular speed, in rev/min, would it have to have in order for the centripetal acceleration of a passenger on the outside edge to be 0.80g, where g = 9.8 m/s2?

answer

watching

128

views

For unlimited access to Homework Help, a Homework+ subscription is required.

You have 0 free answers left.
Get unlimited access to 3.8 million step-by-step answers.

Get unlimited access

Already have an account? Log in

Jean KeelingLv2

23 Nov 2019

Unlock all answers

Get 1 free homework help answer.

Already have an account? Log in

Related questions

An amusement park ride consists of a rotating circular platform11.6 m in diameter from which 10 kg seats are suspended at the endof 3.39 m massless chains. When the system rotates, the chains makean angle of 41.1? with the vertical.
The acceleration of gravity is 9.8 m/s2 .

What is the speed of each seat? Answer in units of m/s.

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 7500 rev/min.

A.) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.

B.) What is the linear speed in m/s of a point on its edge (in m/s)?

sapphirebadger799

In a popular amusement park ride, a rotating cylinder of radius 3.00 m is set in rotation at an angular speed of 5.00 rad/s, as in Figure P7.73. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? Hint: Recall that the magnitude of the maximum force of static friction is equal to , where n is the normal force--in this case, the force causing the centripetal acceleration.