Physics 2053C – Fall 2001

1. Physics 2053C – Fall 2001 Chapter 5 Circular Motion & Gravitation Dr. Larry Dennis, FSU Department of Physics

2. Forces • Newton’s Laws: • If the force on an object is zero, then it’s velocity is constant. • The acceleration = net force / mass. • Whenever one object exerts a force on a second object, the second object exerts an equal and opposite force on the first object. F = m a

3. W = Mg r Tt M Vertical Circular Motion • Consider the system below in which the mass M (= 0.8 kg) is moving in a vertical circle with a speed v. If v = 3.4 m/s and r = 0.65 m. F = Ma Tt + W = Ma Free body diagram for the mass when it is at the top of the circle.

4. Tb r W = Mg M Vertical Circular Motion • Consider the system below in which the mass M (= 0.8 kg) is moving in a vertical circle with a speed v. If v = 3.4 m/s and r = 0.65 m. F = Ma Tb – W = Ma Free body diagram for the mass when it is at the bottom of the circle.

5. r M Vertical Circular Motion • Consider the system below in which the mass M (= 0.8 kg) is moving in a vertical circle with a speed v. If v = 3.4 m/s and r = 0.65 m. Tb – W = Ma Tb = Ma + W Tt + W = Ma Tt = Ma - W Tb - Tt = Ma + W – ( Ma – W ) Tb - Tt = 2W = 2Mg

6. r M Vertical Circular Motion • Consider the system below in which the mass M (= 0.8 kg) is moving in a vertical circle with a speed v. If v = 3.4 m/s and r = 0.65 m. Tb = Ma + W Tb = 0.8 kg * (3.4 m/s)2/ 0.65 m + 0.8 kg*10 m/s2 Tb = 22.2 N

7. r M Vertical Circular Motion • Consider the system below in which the mass M (= 0.8 kg) is moving in a vertical circle with a speed v. If v = 3.4 m/s and r = 0.65 m. Tt = Ma - W Tt = 0.8 kg * (3.4 m/s)2/ 0.65 m – 0.8 kg*10 m/s 2 Tt = 6.2 N

8. Motion in a Horizontal Circle M • Consider the system at the right in which the mass M is moving in a circle with a speed v. How is the radius of the circle related to the mass m? • The small mass puts a tension on the cord that along with the velocity, determines the radius of the circle. Side View r m

9. M Side View r N T m Mass M: Mg Motion in a Horizontal Circle • Draw free body diagrams for masses M and m. Mass m: T mg

10. M Side View r m Mass m: T mg Motion in a Horizontal Circle • Apply Newton’s 2nd Law to the mass m. T – mg = ma = 0 So: T = mg Note this is the same T that is applied to the mass M.

11. M Side View r N T m Mass M: Mg Motion in a Horizontal Circle • Apply Newton’s 2nd Law to the mass M. Vertical Direction N – Mg = Ma = 0 Horizontal Direction T = Ma T = M v2/r Note: a = v2/r because this is uniform circular motion.

12. M Side View r m T = M v2/r and T = mg Motion in a Horizontal Circle • Combine the equations for the masses M and m. So: M v2/r = mg  r = (M/m) v2/g

13. M Side View r m T = M v2/r and T = mg Motion in a Horizontal Circle Suppose: M = 300 gm, m = 50 gm and the period of rotation is 1.4 s. • What is r? • What is T?. So: M v2/r = mg  r = (M/m) v2/g • v = 2r/ • r = (M/m)v2/g = (M/m) (2r/ )2/g • r = (M/m) (2)2 /(g2) r2

14. M Side View r m Motion in a Horizontal Circle Suppose: M = 300 gm, m = 50 gm and the period of rotation is 1.4 s. • What is r? • What is T?. We need to solve for r. r = (M/m) (2)2 /(g2) r2 Or: r = (m/M) (g2)/(2)2 r = (0.05 kg / .3 kg ) * (10 m/s2 * (1.4 s)2)/(2)2 r = 0.083 m And the speed is 0.37 m/s

15. M Side View r m Motion in a Horizontal Circle Suppose: M = 300 gm, m = 50 gm and the period of rotation is 1.4 s. • What is T?. T = mg T = 0.05 kg * 10 m/s2 T = 0.5 m/s2

16. N W = mg Motion on a Banked Track – CAPA #7 • A circular automobile track is banked at an angle  such that no friction between the road and tires is required when a car travels at 25 m/s. If the radius of the track is 375 m, determine .

17. N   W = mg Motion on a Banked Track – CAPA #7 • A circular automobile track is banked at an angle  such that no friction between the road and tires is required when a car travels at 25 m/s. If the radius of the track is 375 m, determine . Nsin Horizontal Direction Nsin = ma = m v2/R Ncos Vertical Direction Ncos - mg = 0

18. N Nsin  Ncos  W = mg Motion on a Banked Track – CAPA #7 • A circular automobile track is banked at an angle  such that no friction between the road and tires is required when a car travels at 25 m/s. If the radius of the track is 375 m, determine . Vertical Direction Ncos - mg = 0  N = mg/cos Horizontal Direction Nsin = ma  mgsin = m v2/R cos g tan  = v2/R

19. N Nsin  Ncos  W = mg Motion on a Banked Track – CAPA #7 • A circular automobile track is banked at an angle  such that no friction between the road and tires is required when a car travels at 25 m/s. If the radius of the track is 375 m, determine . g tan  = v2/R tan  = v2/(Rg) tan  = (25 m/s)2/(375 m * 9.8 m/s2) tan  = 0.17   = 9.6o

20. Newton’s Law of Gravition • Assumed gravity was the same everywhere in the solar system. F = G M1 M2 R2 This is an attractive force. M1 M2 R

21. Newton’s Law of Gravition • It is a very weak force and perhaps one of the least understood. F = G M1 M2 R2 This is an attractive force. M1 F M2 F

22. Next Time • Chapter 5 – Gravitation. • Look at homework # 3. • Please see me with any questions or comments. See you Friday.