Midterm 3 Ans

# Midterm 3 Ans - PHY302 K. Solutions for mid-term test #3....

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Unformatted text preview: PHY302 K. Solutions for mid-term test #3. Problem # 1 : A rider in the barrel of fun appears motionless relative to the rotating barrel indeed, the whole point of this ride to stay stuck to the wall when the floor drops out but relative to the inertial frame of the ground, the rider moves in a horizontal circle of radius R around the barrels axis. This circular motion has a centripetal acceleration a c = v 2 R = 2 R (1) in a horizontal direction towards the barrels axis. The horizontal force providing for this acceler- ation is the normal force N = ma c = m 2 R (2) from the barrels wall on the riders back. This normal force allows for a static friction force vector F between the rider and the wall; when the floor drops out, it is this friction force with balances the riders weight mg and prevents him/her from falling down. To counteract the weight, the friction force vector F must have direction straight up and magnitude mg . The upward direction is OK because the wall is vertical, but the magnitude is limited by the static friction rule | vector F | F max = s N. (3) Hence, to keep the rider from falling down we need a large enough normal force, mg s N = N mg s . (4) But the normal force is governed by the centripetal acceleration according to eq. (2), so we need m 2 R mg s (5) Strictly speaking, the radius of this circle is a few inches shorter than the barrels radius, the difference being the distance between the riders center of mass and the barrels wall. But I am neglecting this difference because its much smaller than the 5-meter radius of the barrel. 1 and consequently 2 g s R . (6) Note that the riders mass drops out from this inequality. In other words, when the floor drops out, the barrels angular velocity must be at least min = radicalbigg g s R = radicalbigg 9 . 8 m / s . 49 5 . 0 m = 2 . 0 s 1 2 . 0 rad / s . (7) Or in terms of the frequency f = / 2 , the minimal spin rate of the barrel must be 2 . 2 = 0 . 32 rev/s or 19 rev/min. Problem # 2 : The angular velocity and hence the speed and the orbital period of a satellite in a circular orbit of radius R around a planet of mass M follows from fact that the centripetal acceleration a c is provided solely by the Newtonian gravity force. Thus, m a c = F grav = m 2 R = GMm R 2 (8) and hence 2 = GM R 3 , (9) regardless of the satellites own mass m . In terms of the orbital period, T = 2 = 2 radicalbigg R 3 GM . (10) For a satellite in an elliptic orbit, the analysis is more complicated and requires calculus....
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## This note was uploaded on 11/11/2010 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas at Austin.

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Midterm 3 Ans - PHY302 K. Solutions for mid-term test #3....

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