PHYS 1300 Lecture Notes - Lecture 12: Smax, Centripetal Force, Resultant Force

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2 Nov 2017

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Phys 1300 lecture 12 newton"s 2nd law in circular motion. Examples of newton"s 2nd law in circular motion. An object moves at a constant speed ac in a circle of radius r. This object experiences centripetal acceleration: therefore there is a resultant force acting to cause this acceleration. This resultant force is sometimes called the centripetal force. T = 9. 5n: a 1. 0 x 103 kg car moving at a constant speed of 21 m/s crosses the rounded top of a hill that has a radius of curvature of 1. 0 x 102 m (ignore friction). Determine: the normal force exerted on the car. Fy = mac mac = fg fn. Fg: the car"s speed when fn is zero. 0 = mg mv2/r mv2/r = mg v2 = rg v = (cid:3493)(100)(9. 8/^2) v = 31 m/s. It is convention to refer to the sum of the forces acting towards the centre of the circle as the centripetal force (fc).

Related Questions

1. A mass, M = 200 g, that is attached to a string, moves in ahorizontal circular path with radius r = 0.5 m. If the speed of themass is 2 m/s, what is the centripetal acceleration? What is thetension in the string?
2. How long does it take the mass to complete one revolution? Howmany revolutions does the mass complete in 2 min?
3. A tetherball of mass m = 500 g is suspended from a string oflength L = 1.5 m. The ball revolves in a horizontal circle movingwith constant speed. The rope makes an angle of 300 with thevertical during the motion. Find the speed of the tetherball.

use this chart for reference

1. An object of mass 19 kg is placed on incline with friction.The incline is originally horizontal and then raised slowly and at21 degrees, the mass begins to slide down the incline. It is foundthat the coefficient of kinetic friction is 0.79 times thecoefficient of static friction. If the same object (same mass) isplaced on the same incline with an inclination angle of 42.2degrees (the mass will slide down the incline at this angle), whatis the magnitude of the acceleration on the mass in m/s2?
2. A mass enters the bottom of a 50 degree incline with acoefficient of kinetic friction of 0.6. At the bottom of theincline the mass is moving at 33 m/s and is moving up the incline.After the mass has traveled, 25 meters up the incline, what is themagnitude of its velocity in m/s?
3. A car is traveling around a circular, banked track withoutfriction. It is initially traveling around the bottom of the ramp(on the ground) where the inner radius of track is 351 meters. Thecar then increases its velocity by a factor of 1.7. If theinclination angle is 10 degrees, what is the height (verticaldistance above the ground) the car is traveling at this newvelocity. Answer in meters.
4. An yo-yo with a mass of 2.4 kg is swung in a vertical circle. Itis found that the string breaks at 228 Newtons. If its the yo-yo'sspeed is a constant and is the maximum speed the yo-yo can movebefore the string breaks, what is the the tension in the string atthe top of the circle in Newtons?
5. A roller coaster cart of mass 15.8 kg travels on the inside of aroller coaster track in a loop of radius 23.1 meters. The circle isa vertical circle similar to the yo-yo example done in class. Ifthe cart travels at a constant speed around the track and the valueof this speed is the minimum speed such that cart does not fall offthe track at the top of the track, was is the magnitude of thenormal force on the cart at the bottom of the track in Newtons?

PHYS 1300 Lecture Notes - Lecture 12: Smax, Centripetal Force, Resultant Force

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