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Physics 121

Physics 121. 5. Kinematics in 2-D and Vectors. 5.1 Kinematics of Uniform Circular Motion 5.2 Dynamics of Uniform Circular Motion 5.3 A Car Rounding a Curve 5.6 Newton’s Law of Universal Gravitation 5.7 Gravity near the Earth’s Surface 5.8 Satellites and Weightlessness. P. Q.

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Physics 121

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  1. Physics 121

  2. 5.Kinematics in 2-D and Vectors 5.1 Kinematics of Uniform Circular Motion 5.2 Dynamics of Uniform Circular Motion 5.3 A Car Rounding a Curve 5.6 Newton’s Law of Universal Gravitation 5.7 Gravity near the Earth’s Surface 5.8 Satellites and Weightlessness

  3. P Q Example 5.1 . . . Round and round A red car goes counterclockwise around a track at a constant speed (see figure) A. The car is accelerating and Q shows the direction of the force on it B. The car is accelerating and P shows the direction of the force on it C. The car is not accelerating D. The car is accelerating but there is no force acting on it.

  4. P Q Solution 5.1 . . . Round and round A. The car is accelerating and Q shows the direction of the force on it. • Centripetal Force changes the direction • Centripetal Force does not change speed • Centripetal Force points toward the center

  5. Example 5.2 . . . Centripetal Force Equation The correct equation for centripetal force is A. F = mv B. F = mvr C. F = mv / r D. F = mv2 / r

  6. Solution 5.2 . . . Centripetal Force Equation The correct equation for centripetal force is F = mv2 / r

  7. Example 5.3 . . . Centripetal Acceleration The correct equation for centripetal acceleration is A. a = mvr B. a = vr C. a = v / r D. a = v2 / r

  8. Solution 5.3 . . . Centripetal Acceleration The correct equation for centripetal acceleration is D. a = v2 / r

  9. Example 5.4 . . . Stoned and Strung Dennis the menace ties a 250 g rock to a 160 cm. string and whirls it above his head. The string will break if the tension exceeds 90 N. What minimum speed will endanger the windshield on his neighbor’s car?

  10. Solution 5.4 . . . Stoned and Strung Centripetal Force is F = mv2 / r 90 = (0.25)(v2 / 1.6 ) v = 24 m/s

  11. Example 5.5 . . . Slippery when wet! A car exits on a ramp (unbanked) of radius 20 m. The coefficient of friction is 0.6. The maximum speed before slipping starts is most nearly A. 10 m/s B. 20 m/s C. 40 m/s D. 120 m/s

  12. Solution 5.5 . . . Slippery when wet! Force of Friction = Centripetal force (0.6)(m)(g) = (m)(v2) / r v = 11 m/s

  13. Is there gravity on Mars? Newton's Law of Universal Gravitation F = GmM/r2 Compare with F = mg so g = GM/r2 • g depends inversely on the square of the distance • g depends on the mass of the planet • g on the Moon is 1 /6 of g on Earth

  14. Example 5.6 . . . A Neutron Star’s Gravity An exotic finish to massive stars is that of a neutron star, which might have as much as five times the mass of our Sun packed into a sphere about 10 km in radius! Estimate the surface gravity on this monster.

  15. Solution 5.6 . . . A Neutron Star’s Gravity g = GM/r2 g = (6.67x10-11)(5x1.99x1030) / (104)2 g = 6.7x 1012 m/s2

  16. Example 5.7 . . . Satellites Geosynchronous satellites are used for cable TV transmission and weather forecasting. They orbit about 36,000 km above the Earth’s surface. This is six times the radius of the Earth which is 6,000 km. What is the value of “g” at that height?

  17. Solution 5.7 . . . Satellites The distance of the satellite from the center of the Earth is 7 RE so the acceleration due to gravity must be 1 / 49 that on the surface of the Earth. (1/49)(9.8) = 0.2 m/s2

  18. That’s all folks!

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