PH1104/PH114S - MECHANICS

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1 PH04/PH4S - MECHANICS FAISAN DAY FALL 06 MULTIPLE CHOICE ANSWES. (E) the first four options are clearly wrong since v x needs to change its sign at a moment during the motion and there s no way v x could switch at an instant from positive to negative (or vice versa). So we are left with the last option. Or you can think it this way; the glider accelerates with constant acceleration of g sin φ towards downhill. Therefore, the curve of v x t should have a constant gradient.. (C) as long as you remember cardinal directions and know how to apply Phytagorean theorem, it s easy. 3. (D) some amount of U grav will be converted into U el. 4. (A) Apply Newton s law of motion. Fx = ma s = N = mω and Fy = 0 = f smax mg = 0 = µ s N mg = 0 = ω min = g µ s As we can see, the larger ω results in larger N. Note that when ω > ω min, the person starts to slide up inside the cylinder since f s has exceeded mg. So by the definition of static friction, in ω > ω min case, the static friction is just the same as maximum static friction which does not play role anymore. Therefore, they both share the same value as in ω = ω min case. 5. (A) the centre of mass of the system is proportionally closer to the larger mass. 6. (B) the star s moment inertia decreases as its radius decreases since I mr. ecall that the rotational kinetic energy can be written as T = L /I or alternatively T = Lω/. By the first argument, it is obvious that T increases as I decreases. The other way to solve the problem is by knowing the fact that the star s angular momentum (L = Iω) has to be conserved, thus decreasing I will lead the star to increase its ω. Therefore, T will increase as ω increases. 7. (C) Just in case you re not aware, Pluto moves in elliptical orbit instead of circular orbit. ecall Kepler s law of orbits. Thus, the force of gravity in elliptical orbit does work on Pluto since there is a component of the force in the direction Pluto moves. W = C 0 {}}{ F ds = W 0 i

2 ecall Kepler s second law, since Pluto moves much faster at perihelion than it does at aphelion then the work done (W = T = U) will be negative. But note that for any type of path an object moves in gravitational field, the work done in going around that path is zero. It follows from W = Fds = 0 8. (B)The pulley will rotate anti-clockwise since m > m. Analyse the torque on pulley s centre, By rearranging the expression above, we have t = τ i = Iα i T T = Iα (m m ) g = I (ω ω 0) t Iω (m m ) g seconds 9. (A) the hint in the question is the key to solve this problem. From the hint given, it s easy to see that I/M < I /M thus IM /I M <. By comparing the rotational kinetic energy, we have By the magical algebra, we have K K = K T K K K T Iω Iv = I v = IK T /M I K T /M I ω = I/M I /M = K T K < K T K 0. (B) the sum of torques (on O) should be zero for the bar to be in equlibrium. i τ i = 0 = F l mg sin θ l = 0 = F = mg sin θ ii

3 SOLUTION # (a) Let s gather all quantities given in the question. = H = 400 m, v 0 = 50 m/s, φ = 30, φ = 30, D = 00 m, L = 00 m So basically we are just required to find the horizontal distance at the moment the boulder touches the ground. Solving the quadratic equation, we have y = y 0 + v 0y t gt = 0 = H v 0 sin φ t gt t = v 0 sin φ + v0 sin φ + gh g Thus the maximum horizontal distance χ reached by the boulder is It turns out that she was right. 6.8 seconds [ v0 sin φ + v0 χ = v 0x t = v 0 cos φ sin φ + gh g ] 96 meter At the moment it touches the ground, its velocity components are v x = v 0x = v x = 5 3 m/s v y = v 0y gt = v y 4 m/s Then the boulder s speed at the moment it hits the pond is (5 3 ) + ( 4) 60 m/s. (b) On the smaller block: Fx = m a = f = m a = a = µ s g For the smaller block not to slide on the bigger block, they must have the same acceleration. Fx = m a = F f = m a = F max = (m + m ) gµ s = 7.64 N If F exceeds the value above, the smaller block will move with acceleration a = µ k g =.96 m/s. While the bigger block will move with acceleration a = F m a m = 7.84 m/s iii

4 If F does not exceed the maximum value, they will move with the same acceleration. m a F {}}{ max f = m a = a = F max (m + m ) = f = m F max (m + m ).94 N So the friction force acting on the smaller block will be.94 N while the friction force acting on the bigger block is just the opposite sign of the one acting on the smaller block. (c) By conservation of energy By conservation of momentum mgl ( cos φ) = mv = v = gl ( cos φ).5 m/s m p v p = m b v b + m p v p m p v = (m b m p ) v m b = m pv gh + m p By working with the magical algebra, we arrived with Putting all the values, we obtain m b 4 kg. ) cos φ m b = m p ( + cos ϕ SOLUTION # (a) iv

5 For the motorcycle not to fall, the torque about the centre of gravity must be zero, which means that the vector force F exerted by the ground must have a line of action passing through the center of gravity. Thus tan θ = F c N = mv / = v mg g ( ) v = θ = tan g If we take µ s account, we will have tan θ m = F c + f s N ( ) = v v g + µ s = θ m = tan g + µ s which is greater than the previous case. This makes sense since f s tends to prevent the motorcycle to slip i.e. it increases the lean angle at which motorcycle could reach to execute the turn perfectly. (b) Analyse the forces acting on the car. and Analyse the total torque on the car s centre of mass. Fy = 0 {}}{ ma y N a + N b = mg (3.) f a + f b = mω + mω ( + l) = mω ( + l) (3.) τcom = 0 = N A l + f A h + f b h N B l = 0 = N B N A = mω ( + l) h l (3.3) v

6 Eliminating (3.3) and (3.) gives = N A = ( mg mω h + ) l = N B = ( mg + mω h + ) l As we can see, the normal force that will vanish is N A with gl ω c = h (l + ) and of course wheel B will lost touch with the ground. In the case ω > ω c, the torque on the centre of mass should be zero for the car to not roll over. τ = 0 N A sin θ l + h + f A cos θ l + h = 0 Note that for ω > ω c, N A will be negative and the outer wheels can t oppose the moment of centrifugal force anymore. Therefore, a car is more likely to roll over around a turn. SOLUTION #3 (a) The angular velocity of the Earth s rotation about its axis is ω = π T = π s rad/s. By Kepler s third law and considering satelite s mass is way much smaller than that of Earth s, we have The change in gravitational potential energy is T ( + H) 3 4π GM E = H meter U = U U = GM Em + H + GM Em = mgh + H 650 megajoules The change in kinetic energy is just the opposite sign of the change in gravitational energy, that is 650 megajoules. (b) Let s denote mass of the Sun as M, the Earth s as M E, and the sattelite s as m. Now analyse the Sun. F g = F c GM M E = M ω ( com ) M E ω com = M ω ( com ) vi

7 We obtain com = M M + M E Now analyse the satellite. F g = F c GM m ( r ) GM Em r = mω ( com r ) From the previous result, we know that com = M E M +M E therefore ω = G(M +M E ) GM. 3 3 By considering r and M M E m, the above expression becomes GM By dividing both sides by M E, we have 3M r M E 3 r ( + r ) GM E r 3M r 3 M E r 0 = r 3 = GM [ ( 3 M ) ] E r M = M E M E 3M = r meter vii