RCW4 solutions - Recitation ClassWork 4 solutions 1....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Recitation ClassWork 4 solutions Physics 203 1. Chapter 5 problem 5 REASONING The magnitude a c of the car’s centripetal acceleration is given by Equation 5.2 as 2 c / a vr = , where v is the speed of the car and r is the radius of the track. The radius is r = 2.6 × 10 3 m. The speed can be obtained from Equation 5.1 as the circumference (2 π r ) of the track divided by the period T of the motion. The period is the time for the car to go once around the track ( T = 360 s). SOLUTION Since 2 c / = and ( ) 2/ v rT = , the magnitude of the car’s centripetal acceleration is ( ) ( ) 2 23 22 2 2 c 2 2 4 2.6 10 m 4 0.79 m/s 360 s r vr T a rr T   ×  = = = = = 2. Problem 11 The sample makes one revolution in time T as given by T = 2 r / v . The speed is v 2 = ra c = (5.00 × 10 –2 m)(6.25 × 10 3 )(9.80 m/s 2 ) so that v = 55.3 m/s The period is T = 2 (5.00 × 10 –2 m)/(55.3 m/s) = 5.68 × 10 –3 s = 9.47 × 10 –5 min The number of revolutions per minute = 1/ T = 10 600 rev/min . 3. Problem 22. Before you try to answer this problem mathematically, first draw a force diagram. Once you and your group think you have the diagram right, call your TA over and discuss it with them. If you’re TA isn’t available because they are working with another group, find another group in the class to discuss your ideas with. REASONING The coefficient µ s of static friction is related to the magnitude MAX s f of the maximum static frictional force and the magnitude F N of the normal force acting on the car by MAX s sN fF = (Equation 4.7), so that: MAX s s N f F = (1) The car is going around an unbanked curve, so the centripetal force 2 c mv F r = (Equation 5.3) must be horizontal. The static frictional force is the only horizontal force, so it serves as the centripetal force. The maximum centripetal force occurs when MAX cs Ff = . Therefore, the maximum speed v the car can have without slipping is related to MAX s f by 2 MAX mv r = (2) Substituting Equation (2) into Equation (1) yields
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 s N
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

RCW4 solutions - Recitation ClassWork 4 solutions 1....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online