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An85.0kgpassenger is made to move along a circular path of radiusr=3.50min uniform circular motion. (a) Figure 6-40a is a plot of the required magnitudeFof the net centripetal force for a range of possible values of the passenger’s speed v. What is the plot’s slope atV=8.30m/s? (b) Figure 6-40b is a plot of F for a range of possible values ofT, the period of the motion. What is the plot’s slope atrole="math" localid="1654172716493" T=2.50s?

Short Answer

Expert verified

a. The plot’s slope is403N.s/m

b. The period of the motion and the plot’s slope atT=2.50sis T=2Ï€¸é/Vand localid="1654173227131" -1.50×103N/s

Step by step solution

01

Given

Circular path of radiusr=3.50m

Mass of passenger m= 85kg

02

Determining the concept

This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves centripetal force which makes an object follow the circular path.

Formula:

The velocity in uniform circular motion is given by

v=2Ï€°ùT (i)

where, vis the velocity, r is the radius andTis the time period

the centripetal force is given by,

F=mv2r (ii)

03

(a) Determining the plot’s slope at 

The slope of the plot at v=8.30m/sis.

dFdv=2mvr=2(85.0kg)(8.30m/s)3.50m=403N.s/m

Hence, the plot's slope is403N.s/m.

04

(b) Determining the period of the motion and the plot’s slope at

The period of the circular ride is

T=2Ï€°ùvThus,F=mv2r=mr2Ï€°ùT2=4Ï€2mrT2

The variation of F with respect to T while holding r constant is,

dF=-8Ï€2mrT3dTTheslopeoftheplotatT=2.50sis.dFdT=-8Ï€2mrT3=-8Ï€2(85.0kg)(3.50m)(2.50s)3=-1.50×103N/sHence,theperiodofthemotionandtheplot'sslopeatT=2.50sisT=2Ï€°ù/vand-1.50×103N/s

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