PHY 302 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 barrel s axis. This circular motion has a centripetal acceleration a c v2 R ω2 R (1) in a horizontal direction towards the barrel s axis. The horizontal force providing for this acceleration is the normal force N ma c mω 2 R (2) from the barrel s wall on the rider s back. This normal force allows for a static friction force F between the rider and the wall; when the floor drops out, it is this friction force with balances the rider s weight mg and prevents him/her from falling down. To counteract the weight, the friction force 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 F F max µ s N. (3) Hence, to keep the rider from falling down we need a large enough normal force, mg µ s N N mg µ s. (4) But the normal force is governed by the centripetal acceleration according to eq. (2), so we need mω 2 R mg µ s (5) Strictly speaking, the radius of this circle is a few inches shorter than the barrel s radius, the difference being the distance between the rider s center of mass and the barrel s wall. But I am neglecting this difference because it s much smaller than the 5-meter radius of the barrel. 1
and consequently ω 2 g µ s R. (6) Note that the rider s mass drops out from this inequality. In other words, when the floor drops out, the barrel s angular velocity ω must be at least g ω min µ s R 9.8 m/s 0.49 5.0 m 2.0 s 1 2.0 rad/s. (7) Or in terms of the frequency f ω/2π, the minimal spin rate of the barrel must be 2.0 2π or 19 rev/min. 0.32 rev/s 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 mω 2 R GMm R 2 (8) and hence ω 2 GM R 3, (9) regardless of the satellite s own mass m. In terms of the orbital period, T 2π ω 2π R 3 GM. (10) For a satellite in an elliptic orbit, the analysis is more complicated and requires calculus. However, the end result for the period is exactly as in eq. (10), except that the radius R should 2
be replaced with the semi-major axis a of the elliptic orbit, T 2π a 3 GM. (11) Now let s compare eqs. (11) for Triton and Luna. Denoting T T and T L their respective orbital periods, and a T and a L the semi-major axes of their orbits, we have T T a 3 T 2π, GM N a 3 L T L 2π. (12) GM E where M N is the Neptune s mass and M E is the mass of the Earth. Taking the ratio of eqs. (12), we obtain and consequently T T T L a 3 T GM N / a 3 L GM E a 3 / T GM N a 3 L GM E a 3 T a 3 L / MN M E ( TT T L ) 2 ( at a L ) 3 / MN M E. (13) From this equation, we find the ratio of the two planet s masses as M N M E ( at a L ) 3 / ( TT T L ) 2 ( ) 3 / ( 354, 800 km 5.877 days 384, 400 km 27.32 days 17.0, ) 2 (14) i.e., Neptune is 17 times more massive than Earth. PS: This is not required for this test and the students who finish with eq. (14) will get full credit, but it s good to know. The actual mass ratio is 17.2, slightly larger than in eq. (14), because Luna s mass does have a a small effect on its orbital motion. 3
While most moons have mass less than 1/1000 of the planet they orbit, the Earth Luna system has unusually small mass ratio M E /M L 81. (In the solar system, only the Pluto Charon system has a smaller ratio.) Consequently, it becomes noticeable that Luna orbits not the Earth s center but the common center of mass of the Earth Luna system, which is about 4700 km closer to the Luna. Hence, in eq. (8), the radius R in the formula for centripetal acceleration is not quite the same as the distance in the formula for the gravity force. Re-deriving the orbital equations to account for this effect, one ends up with a 3 L T L 2π G(M E + M L ) instead of eq. (11). Similar corrections apply to any binary system in which the satellite s mass cannot be neglected compared to the mass of its primary. Therefore, eq. (14) should be modified as (15) M N + M T M E + M L ( at ) 3 / ( ) 2 TT 17.0. (16) a L T L In the numerator on the left hand side, we may neglect Triton s mass M T because it s almost 5000 times smaller than Neptune s. But Luna s mass is more noticeable compared to Earth s, so we should say that Neptune is 17 times more massive than Earth and Luna together, or about 17.2 times the mass of Earth alone. Problem # 3: There are three force on the meterstick: tension T of the upper string, tension T mg (where m 50 g) of the lower string, and its own weight Mg. Here is the force diagram: T CM 10 40 50 T mg Mg Although the weight force Mg is distributed all over the meterstick, for the purpose of calculating 4
torques, we treat it as acting at the center of mass, and that s what the diagram shows. By symmetry, the center of mass is in the middle of the meterstick, at the 50 cm mark. The meterstick is in equilibrium, so the net force and the net torque on it must be zero, F 0, τ 0. (17) In the torque condition, we may calculate the torques relative to any pivot point we like (as long as it s the same point for all the forces), so let s consider the net torque relative to the 40 cm mark where the upper string is attached. With this choice, the tension T of the upper string has zero lever arm, the tension T mg of the lower string has lever arm 40 cm 10 cm 30 cm in the counterclockwise direction, and the meterstick s own weight Mg has lever arm 50 cm 40 cm 10 cm in the clockwise direction. Consequently, the net torque is τ net τ(t) + τ(mg) + τ(mg) 0 mg 30cm + Mg 10cm. (18) Demanding that this net torque vanishes, we obtain Mg 10cm mg 30cm 0, (19) and consequently the meterstick s mass is M m 30 cm 10 cm m 3 50 g 3 150 g. (20) Problem # 4: The wheel s angular acceleration or rather deceleration follows from the net torque on the wheel according to Iα τ net (21) The force acting on the wheel include the normal force N from the rag, the kinetic friction force F, and also some unknown force F A at the axle of the wheel. (The F A keeps the axle from moving by canceling N + F.) Fortunately, regardless of the directing and magnitude of the axle force F A, it has zero lever arm with respect to the wheel s axis of rotation and thus does not produce a torque, τ( F A ) 0. 5
The normal force N acts at the wheel s tire, at distance R from the axis, but the direction of this force is radial, so the line of force goes right through the axis. Consequently, N also has a zero lever arm and produces no torque, τ( N) 0. Finally, the friction force F acts at the place as N, at distance R from the axis of rotation, but the direction of F is tangent to the tire and perpendicular to the radius vector. Consequently, the lever arm of F is l R sin 90 R, and the torque is τ( F) F R. Or rather, τ( F) F R because this torque has opposite direction to the wheel s rotation. Altogether, the net torque on the wheel is τ net τ( F A ) + τ( N) + τ( F) 0 + 0 F R. (22) Hence, according to eq. (21), the angular acceleration of the wheel is α τnet I F R M R 2 F M R 15 N 3.0 kg 0.33 m 15 s 2 15 rad/s 2. (23) Now consider the kinematics of the wheel s rotation. Since the angular acceleration (23) is constant, we have Eq. (24) gives us the time it takes the wheel to stop, ω(t) ω 0 + αt ω 0 α t, (24) ϕ(t) ω 0 t + 1 2 αt2 ω 0 t 1 2 α t2. (25) ω 0 α t 0 t ω 0 α, (26) and then eq. (25) gives us the angle through which the wheel rotates while stopping, ϕ ω 0 ω 0 α α 2 ( ) 2 ω0 ω2 0 α α ω2 0 2 α ω2 0 2 α (60 rad/s)2 120 rad. (27) 2 15 rad/s2 For each 2π radians of this angle, the wheel makes one full revolution. So the number of revolutions of the wheel before it stops is #rev ϕ 2π 19. (28) 6
Problem # 5: The water in a lake is at rest, so the hydrostatic equation tells us that the pressure increases with depth as P const + ρg depth (29) where ρ 1000.0 kg/m 3 is the density of water. At the surface of the lake, the pressure equals to the atmospheric pressure, P sur P atm 1.013 10 5 Pa, (30) so at the bottom of the lake the pressure is P bot P atm + ρgd (31) where D is the lake s depth. Note that both pressures (30) and (31) are absolute pressures rather than gauge pressures. According to the Boyle s Law volume of a fixed amount of gas held at constant temperature is inversely proportional to the absolute pressure of the gas, hence P V const. For the bubble in question, this means P bot V @bot P sur V @sur (32) and consequently P bot P sur V @sur V @bot 3.0 cm3 3.0. (33) 1.0 cm3 On the other hand, according to eqs. (30) and (31), P bot P sur P atm + ρgd P atm 1 + ρgd, (34) P atm Historically, several gas laws were discovered between 17 th and 19 th centuries. The oldest gas law relating the pressure and the volume of a gas was discovered in 1664 by Robert Boyle in Ireland, hence the name Boyle s Law. (The same law was independently discovered in 1676 by Edme Mariotte in France, so in some countries this law is called the Boyle Mariotte Law.) Eventually, Boyle s law was combined with Charles s, Gay Lussac s, and Avogadro s laws into the universal gas law (also known as the ideal gas law). If you have taken a chemistry class, you should already know this law in all its aspects. If not, you can look it up in 10.4 of the Serway and Faughn textbook.) 7
and combining this formula with eq. (33), we arrive at This is a simple equation for the lake s depth D, and the solution is 1 + ρgd 3.0. (35) P atm D P atm ρg (3.0 1) 101300 N/m 2 1000.0 kg/m 3 2.0 21 m. (36) 9.8 N/kg 8