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Chapter 8: Q. 64 (page 202)

In Problems 64 and 65 you are given the equation used to solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation. Besure that the answer your problem requests is consistent with the equation given.
b. Finish the solution of the problem.
60 N = (0.30 kg)ω2(0.50 m)

Short Answer

Expert verified

a) A ball of mass 300 gm , attached to a string , is rotating around a horizontal circle of 0.5 m radius. The tension on the string is 60 N . Find the angular velocity of the ball.

b) Angular velocity is 20 rad/sec or 190.985 rmp.

Step by step solution

01

Part(a) Step 1 : Given information

The equation given is 60 N = (0.30 kg)ω2 (0.50 m)

02

Part(a) Step 2 : Explanation

Lets observe the equation given 60 N = (0.30 kg)ω2 (0.50 m)

We have Force =60 N. Mass 0.3 kg and radius 0.5 m. Angular velocity is unknown.

We can state problem as

A ball of mass 300 gm , attached to a string , is rotating around a horizontal circle of 0.5 m radius. The tension on the string is 60 N . Find the angular velocity of the ball.

03

Part(b) Step 1 : Given information

The equation given is 60 N = (0.30 kg)ω2 (0.50 m)

04

Part(b) Step 2 : Explanation

The centripetal force acting on a mass m revolving with angular speed ωaround a circle of radius r is given by mω2 r.

The angular velocity of the ball is calculated as

60N=(0.30kg)ω2(0.50m)ω2=60N(0.30kg)(0.50m)ω=20rad/s

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In Problems 64 and 65 you are given the equation used to solve a problem. For each of these, you are to
a. Write a realistic problem for which this is the correct equation. Be sure that the answer your problem requests is consistent with the equation given.
b. Finish the solution of the problem.
65. (1500 kg)(9.8 m/s2) - 11,760 N = (1500 kg) v2/(200 m)
See all solutions

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