176_PartUniversity Physics Solution

# 176_PartUniversity Physics Solution - 5-48 Chapter 5(c For...

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5-48 Chapter 5 (c) For angle 2 r , sin cos 0 mg mg Dv ββ μβ −− = and r (sin cos ) mg v D βμ β = . The terminal speed for a falling object is derived from 2 t 0, Dv mg −= so t . vm g D = tr sin cos vv =− . And since rt 0.015, sin (0.015) cos == . EVALUATE: In part (c), t as 90 ° , since in that limit the incline becomes vertical. 5.108. IDENTIFY: Apply m = F a ! ! to the person and to the cart. SET UP: The apparent weight, app w , which is the same as the upward force on the person exerted by the car seat. EXECUTE: (a) The apparent weight is the actual weight of the person minus the centripetal force needed to keep him moving in its circular path : 22 2 app (12 m s) (70 kg) (9.8 m s ) 434 N 40 m mv wm g R ⎡⎤ = = ⎢⎥ ⎣⎦ . (b) The cart will lose contact with the surface when its apparent weight is zero; i.e., when the road no longer has to exert any upward force on it : 2 0 mv mg R . 2 (40 m) (9.8 m/s ) 19.8 m s vR g = . The answer doesn&t depend on the cart&s mass, because the centripetal force needed to hold it on the road is proportional to its mass and so to its weight, which provides the centripetal force in this situation. EVALUATE: At the speed calculated in part (b), the downward force needed for circular motion is provided by gravity. For speeds greater than this more, downward force is needed and there is no source for it and the cart leaves the circular path. For speeds less than this, less downward force than gravity is needed, so the roadway must exert an upward vertical force. 5.109. (a) IDENTIFY: Use the information given about Jena to find the time t for one revolution of the merry-go-round. Her acceleration is rad , a directed in toward the axis. Let 1 F ! be the horizontal force that keeps her from sliding off. Let her speed be 1 v and let 1 R be her distance from the axis. Apply m = F a ! ! to Jena, who moves in uniform circular motion. SET UP: The free-body diagram for Jena is sketched in Figure 5.109a EXECUTE: x x Fm a = 1r a d a = 2 1 1 1 , v R = 11 1 1.90 m/s RF v m Figure 5.109a The time for one revolution is 1 1 1 2 2. R m tR F π Jackie goes around once in the same time but her speed 2 () v and the radius of her circular path 2 R are different. 21 1 2 1 1 2 R R m R m ⎛⎞ = ⎜⎟ ⎝⎠ IDENTIFY: Now apply m = F a ! ! to Jackie. She also moves in uniform circular motion. SET UP: The free-body diagram for Jackie is sketched in Figure 5.109b.

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176_PartUniversity Physics Solution - 5-48 Chapter 5(c For...

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