A puck of mass m=1.50kgslides in a circle of radiusr=20.0cmon a frictionless table while attached to a hanging cylinder of massM=2.50kgby means of a cord that extends through a hole in the table (Fig. 6-43). What speed keeps the cylinder at rest?

Short Answer

Expert verified

The speed that keeps the cylinder at rest is 1.81 m/s.

Step by step solution

01

Given information

  • Mass of pluck,m=1.50kg.
  • Mass of hanging cylinder,M=2.50kg.
  • The radius of the circle, r=20.0cm.
02

To understand the concept

The problem deals with the centripetal force. It is a force that makes a body follow a curved path.For the puck to remain at rest, the magnitude of the tension forceTof the cord must equal the gravitational forceMgon the cylinder. The tension force supplies the centripetal force that keeps the puck in its circular orbit.

03

To calculate the speed that keeps the cylinder at rest

In the equilibirum condition:

Tensionforce=CentripetalforceT=mv2/r

But,

T=Mg

Thus,

Mg=mv2/r

We solve for the speed as:

v=Mgrm

Substitute the values, and we get,

v=(2.50 kg)(9.80 m/s2)(0.200 m)1.50 kgv=1.81 m/s

Thus, the speed that keeps the cylinder at rest is 1.81 m/s

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Most popular questions from this chapter

A student wants to determine the coefficients of static friction and kinetic friction between a box and a plank. She places the box on the plank and gradually raises one end of the plank. When the angle of inclination with the horizontal reaches30°, the box starts to slip, and it then slides 2.5 mdown the plank in 4.0 sat constant acceleration. What are

(a) the coefficient of static friction and

(b) the coefficient of kinetic friction between the box and the plank?

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