151_PartUniversity Physics Solution

# 151_PartUniversity - Applying Newtons Laws 5-23 The person moves in a circle of radius R = 3.00 m(5.00 m)sin 30.0 = 5.50 m The free-body diagram is

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

Applying Newton&s Laws 5-23 SET UP: The person moves in a circle of radius 3.00 m (5.00 m)sin30.0 5.50 m R =+ = ° . The free-body diagram is given in Figure 5.52. F ! is the force applied to the seat by the rod. EXECUTE: (a) y y Fm a = gives cos30.0 g = ° and cos30.0 mg F = ° . x x a = gives 2 sin30.0 v R = ° . Combining these two equations gives 2 tan (5.50 m)(9.80 m/s )tan30.0 5.58 m/s vR g θ == = ° . Then the period is 22 ( 5 . 5 0 m ) 6.19 s 5.58 m/s R T v ππ = . (b) The net force is proportional to m so in m = F a ! ! the mass divides out and the angle for a given rate of rotation is independent of the mass of the passengers. EVALUATE: The person moves in a horizontal circle so the acceleration is horizontal. The net inward force required for circular motion is produced by a component of the force exerted on the seat by the rod. Figure 5.52 5.53. IDENTIFY: Apply m F =a ! ! to the composite object of the person plus seat. This object moves in a horizontal circle and has acceleration rad a , directed toward the center of the circle. SET UP: The free-body diagram for the composite object is given in Figure 5.53. Let x + be to the right, in the direction of rad a ! . Let y + be upward. The radius of the circular path is 7.50 m R = . The total mass is 2 (255 N 825 N)/(9.80 m/s ) 110.2 kg += . Since the rotation rate is 32.0 rev/min 0.5333 rev/s = , the period T is 1 1.875 s 0.5333 rev/s = . EXECUTE: y y a = gives cos40.0 0 A Tm g = ° and 255 N 825 N 1410 N cos40.0 cos40.0 A mg T + = °° . x x a = gives rad sin40.0 AB TT m a ° and 44 ( 7 . 5 0 m ) sin40.0 (110.2 kg) (1410 N)sin40.0 8370 N (1.875 s) BA R T T =− = = . The tension in the horizontal cable is 8370 N and the tension in the other cable is 1410 N. EVALUATE: The weight of the composite object is 1080 N. The tension in cable A is larger than this since its vertical component must equal the weight. rad 9280 N ma = . The tension in cable B is less than this because part of the required inward force comes from a component of the tension in cable A . Figure 5.53

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

View Full Document
5-24 Chapter 5 5.54. IDENTIFY: Apply m F =a ! ! to the button. The button moves in a circle, so it has acceleration rad a . SET UP: The situation is equivalent to that of Example 5.22. EXECUTE: (a) 2 s v Rg μ = . Expressing v in terms of the period T , 2 R v T π = so 2 s 2 4 R Tg = . A platform speed of 40.0 rev/min corresponds to a period of 1.50 s, so 2 s 22 4 (0.150 m) 0.269.
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

### Page1 / 5

151_PartUniversity - Applying Newtons Laws 5-23 The person moves in a circle of radius R = 3.00 m(5.00 m)sin 30.0 = 5.50 m The free-body diagram is

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

View Full Document
Ask a homework question - tutors are online