Trial Hanging Mass, M Radius, r Moving (changes) Mass, m Moving Experimental (1 large (1 small washer) washer) 1 0.02431 kg 0.30 m 0.00186 mass/ radius m/r 0.0062 Velocity Data Table 6: Calculations: Centripetal Force-Varying the radius of the Moving mass, r Weight (Hanging mass * g) (N) VE² Centripetal Force (N) m(VE)² Mg r 1.37 m/s 3.61 m²/s² 0.081 N 0.239 N kg 2 0.02431 kg 0.25 m 0.00186 0.0074 1.45 m/s 4.0 m²/s² 0.020 N 0.239 N kg 3 0.02431 kg 0.20 m 0.00186 0.0093 1.53 m/s 3.24 m²/s² 0.098 N 0.239 N kg 4 0.02431 kg 0.15 m 0.00186 0.0124 1.64 m/s 5.76 m²/s² 0.411 N 0.239 N kg Centripetal Force (N) 0.4 0.3 2 0.1 0.0- Centripetal Force vs. Experimental Velocity ■plot -linear fit F=0.14039-0.43048 3.0 3.5 4.0 4.5 5.0 5.5 6.0 VE² (m²/s²)

Trial Hanging Mass, M Radius, r Moving (changes) Mass, m Moving Experimental (1 large (1 small washer) washer) 1 0.02431 kg 0.30 m 0.00186 mass/ radius m/r 0.0062 Velocity Data Table 6: Calculations: Centripetal Force-Varying the radius of the Moving mass, r Weight (Hanging mass * g) (N) VE² Centripetal Force (N) m(VE)² Mg r 1.37 m/s 3.61 m²/s² 0.081 N 0.239 N kg 2 0.02431 kg 0.25 m 0.00186 0.0074 1.45 m/s 4.0 m²/s² 0.020 N 0.239 N kg 3 0.02431 kg 0.20 m 0.00186 0.0093 1.53 m/s 3.24 m²/s² 0.098 N 0.239 N kg 4 0.02431 kg 0.15 m 0.00186 0.0124 1.64 m/s 5.76 m²/s² 0.411 N 0.239 N kg Centripetal Force (N) 0.4 0.3 2 0.1 0.0- Centripetal Force vs. Experimental Velocity ■plot -linear fit F=0.14039-0.43048 3.0 3.5 4.0 4.5 5.0 5.5 6.0 VE² (m²/s²)

Classical Dynamics of Particles and Systems

5th Edition

ISBN:9780534408961

Author:Stephen T. Thornton, Jerry B. Marion

Publisher:Stephen T. Thornton, Jerry B. Marion

Chapter9: Dynamics Of A System Of Particles

Section: Chapter Questions

Problem 9.63P

See similar textbooks

Similar questions

The mean diameter of the planet Mercury is 4.88106m , and the acceleration due to gravity at its surface is 3.78m/s2 . Estimate the mass of this planet.
In the law of universal gravitation, Newton assumed that the force was proportional to the product of the two masses (m1m2) . While all scientific conjectures must be experimentally verified, can you provided arguments as to why this must be? (You may wish to consider simple examples in which any other form would lead to contradictory results.)
The acceleration due to gravity on the surface of a planet is three times as large as it is on the surface of Earth. The mass density of the planet is known to be twice that of Earth. What is the radius of this planet in terms of Earth’s radius?
What is the gravitational acceleration close to the surface of a planet with a mass of 2ME and radius of 2RE where ME, and RE are the mass and radius of Earth, respectively? Answer as a multiple of g, the magnitude of the gravitational acceleration near Earths surface. (See Section 7.5.)
Unreasonable Results (a) Based on Kepler's laws and information on the orbital characteristics of the Moon, calculate the orbital radius for an Earth satellite having a period of 1.00 h. (b) What is unreasonable about this result? (c) What is unreasonable or inconsistent about the premise of a 1.00 h orbit?
In your own words, describe the difference between the terms gravitational force and gravitational field.
Imagine that nitrogen and other atmospheric gases were more soluble in water so that the atmosphere of the Earth is entirely absorbed by the oceans. Atmospheric pressure would then be zero, and outer space would start at the planets surface. Would the Earth then have a gravitational field? (a) Yes, and at the surface it would be larger in magnitude than 9.8 N/kg. (b) Yes, and it would be essentially the same as the current value. (c) Yes, and it would be somewhat less than 9.8 N/kg. (d) Yes, and it would be much less than 9.8 N/kg. (e) No, it would not.
A satellite of mass 16.7 kg in geosynchronous orbit at an altitude of 3.58 104 km above the Earths surface remains above the same spot on the Earth. Assume its orbit is circular. Find the magnitude of the gravitational force exerted by the Earth on the satellite. Hint: The answer is not 163 N.
In Problem 50, what is the tangent of the angle for which 5.00 in is the opposite side?
Check Your Understanding Find the angle between forces F1and F3in Example 2.16.
Check Your Understanding Why not use the simpler expression U=mg(y2y1) ? How significant would the error be? (Recall the previous result, in Example 13.4, that the value g at 400 km above the Earth is 8.67m/s2 .)
The Lunar Reconnaissance Orbiter (LRO), with mass m = 1850 kg, maps the surface of the Moon from an orbital altitude of 50.0 km. What are the magnitude and direction of a. the force the LRO experiences due to the Moons gravity and b. the force exerted by the LRO on the Moon?