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Chapter 6: Q6.3-12CQ (page 219)

Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth鈥檚 frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton鈥檚 third law, explain what force stretches the string, identifying its physical origin.

Short Answer

Expert verified

There is a real force acting on the string due to the mass of the mail.

Step by step solution

01

Definition of Centrifugal force

When a mass is rotated, the apparent outward force on it is called centrifugal force.

02

Identification of origin based on centripetal force and Newton’s third law

The mass is moving in such a circular path where the force of friction plays no part, but a net centripetal force acts towards the center.

Also, as the nail is attached to an inextensible string, a net tension force acts on the string which pulls it outwards, and as known according to Newton鈥檚 third law of motion, every action must have an equal and opposite reaction. Therefore, this tension force and the centripetal force both are equal and act in opposite directions to each other. Thus, there is a real force acting on the string due to the mass of the mail.

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Most popular questions from this chapter

Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car鈥檚 speed? What is the direction of the force exerted on you by the car seat?

An automobile with 0.260 m radius tires travels80,000km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components鈥攆riction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system鈥檚 weight).

(a) Show that\(\theta \)(as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway鈥攖hat is,\(\theta = {\tan ^{ - 1}}\,{v^2}/rg\)

(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).

Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle鈥攖he ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a \(100{\rm{ m}}\) radius curve banked at \(15.0^\circ \). (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at \(20.0{\rm{ km}}/{\rm{h}}\)?

As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer.
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