Random Problems

1. Random Problems The following is a random collection of problems. There is no special significance of the appearance or the order of these problems.

2. A grinding wheel accelerates from rest to an angular velocity of 675 rev/min in 1.25 min. Determine the: • angular acceleration in rad/s2 ωi = 0 ωf = 675 rev/min Δt = 1.25 min ωf =ωf + Δt α

3. rad rev 2π rad 60 s 60 s min α = 675 × × / s2 min 1 rev min min 60 s = 0.942 rad/s2 (1.25 min × (1.25 min × )2 ) . (b) angular displacement after 1.25 min Δθ = ωiΔt + ½αΔt2 Δθ = 0 + ½ × 0.942 × Δθ = 2650 rad

4. (c) angular displacement during the second 30.0 s interval. Δθ = ωiΔt + ½αΔt2 Δθ60 – Δθ30 = ½ × 0.942 rad/s2 × (60.0 s)2 - ½ × 0.942 rad/s2 × (30.0 s)2 Δθ60 – Δθ30 = 1270 rad

5. A 1 kg mass is hung from a rope which is fastened to the ceiling. What downward force acts on the 1 kg mass? The weight of the 1 kg mass. According to Newton’s 3rd Law, what is the reaction force? The reaction force is the 1 kg mass pulling straight up on the earth.

6. Why do these forces not cancel out? An action-reaction pair of forces are equal in magnitude, opposite in direction, but act on different objects.

7. rev ω 55 = min A grinding wheel has a constant angular velocity of 55 rev/min. • Determine the angular velocity in rad/s. (b) Determine the angular acceleration in rad/s2. If the angular velocity is constant, then the angular acceleration is 0. 1 min 2π rad 5.76 rad/s × × = rev 60 s

8. A car with an initial velocity of 30.0 m/s accelerates for 8.0 s at the rate of 7.0 m/s2. (a) What is the car’s total displacement? vi = 30.0 m/s a = 7.0 m/s2 Δt = 8.0 s Δx = viΔt + 1/2aΔt2 Δx = 30.0 m/s × 8.0 s + ½ × 7.0 m/s2 × (8.0 s)2 Δx = 464 m

9. What is the car’s displacement during the eighth second? Δx = viΔt + 1/2aΔt2 Δx = 30.0 m/s × 7.0 s + ½ × 7.0 m/s2 × (7.0 s)2 Δx = 382 m Δx = 464 m - 382 m = 82 m

10. A 2.0 kg mass is suspended from a spring. When the mass is set in motion and air resistance is negligible. Will the total energy of the system be conserved? Yes, mechanical energy is always conserved in the presence of conservative (non-dissipative) forces such as gravity and the restoring force within a spring.

11. How many forms of potential energy are present in this scenario? There are two forms, elastic potential (½kx2) and gravitational (mgΔh) potential energy.

12. A net force of 15.0 N acts on an object with a mass of 2.5 kg. • If the object starts from rest, how much work is done in the first three seconds? Fnet = 15.0 N m = 2.5 kg vi = 0 W = F × x × cosθ Fnet = ma

13. Fnet m 15.0 N a 6.0 m/s2 = = = 2.5 kg Δx = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2× (3.0 s)2 . Δx = 27 m W = F × x × cosθ W = 15.0 N × 27 m = 4.0 x 102 J

14. Δx7 = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2× (7.0 s)2 (b) How much work is done during the seventh second? Δx7 = 150 m Δx6 = x0 + viΔt + ½aΔt2 = 0 + ½ × 6.0 m/s2× (6.0 s)2 Δx6 = 110 m W = 15.0 N × 150 m – 15.0 N × 110 m = 6.0 x 102 J

15. A ferrous wheel rotates at 0.25 rad/s. • Determine the number of revolutions in 2.5 min. ω = 0.25 rad/s Δt = 2.5 min ω = Δθ/Δt Δθ = 0.25 rad/s × 2.5 min × 60 s/min × 1 rev/2π rad = 6.0 rev

16. (b) Determine the time needed to stop the ferrous wheel if it decelerates at -0.010 rad/s2. α = -0.010 rad/s2 ωf = ωi + αΔt Δt = 0 – 0.25 rad/s/-0.010 rad/s2 = 25 s

17. A soccer ball with a mass of 5.00 kg rolls at 9.00 m/s. If the soccer ball does not slide along the ground, determine its: • linear kinetic energy m = 5.00 kg v = 9.00 m/s KE = ½mv2 = ½ × 5.00 kg × (9.00 m/s)2 KE = 202 J

18. (b) rotational kinetic energy given that I = 2/5mr2. KE = ½Iω2 = ½ × 2/5 × m × r2 × v2/r2 KE = mv2/5 = 5.00 kg × (9.00 m/s)2/5 = 81.0 J (c) Was it important to be told that the soccer ball did not slip or slide along the way? Yes, because if it had, you would to be told for long the ball slid to compute the KErot.

19. (d) What is the significance of not being given the radius of the soccer ball? The radius is not important because it cancels out in the calculation. This fact is important to keep in mind. Sometimes it is helpful to simplify the calculation before performing the calculation. In this case, you would not have been able to do the problem because of the missing radius.

20. Pete is driving his car eastward at the same time, his brother Repeat, is driving westward. Because they were so excited to see each other, they forgot to break and had a head on collision. Pete and his car had a mass of m and Repeat and his car had a mass of 5m. • If Repeat’s original velocity is v, what was Pete’s velocity before the collision? mP = mP mR = 5mP vP = ? vR = v

21. Δp = 0 pi = pf mPvP + mRvR = mPvP’ + mRvR’ mPvP + 5mPvR = 0 vP = -5vR

22. (b) If during the force of impact, Pete exerted a force of F on Repeat, what force did Repeat exert on Pete? According to Newton’s 3rd Law, for every force there is an equal but opposite force, therefore, Repeat exerted on force, F, on Pete.

23. A 6.50 kg mass is dropped from a height of 0.575 m striking a vertically placed spring. If the elastic constant of the spring is 2.50 x 103 N/m, how much did the spring compress? m = 6.50 kg h = 0.750 m k = 2.50 x 103 N/m ΔE = 0 (PEe + PEg)T = (PEe + PEg)B

24. (0 + mgΔh)T = (½kΔx2 + 0)B x = (2 × 6.50 kg × 9.80 m/s2 × 0.575 m/ (2.50 x 103 N/m))½ x = 0.171m

25. A baseball with a mass of 0.145 kg starts rolling from the top of an incline which is 3.0 m high. If the incline is 12.0 m long, determine the: • linear velocity of the baseball when it reaches the bottom. h = 3.0 m l = 12.0 m m = 0.145 kg

26. ΔE = 0 (PEg + KEt + KEr )T = (PEg + KEt + KEr )B mgΔh + 0 = ½mv2 + ½Iω2 mgΔh = ½mv2 + ½ × 2/5 × m × v2/r2 9.80 m/s2 × 3.0 m = 7/10 × v2 v = 6.5 m/s

27. (b) the angular kinetic energy when it reaches the bottom. ΔE = 0 (PEg + KEt + KEr )T = (PEg + KEt + KEr )B mgΔh + 0 = ½mv2 + ½Iω2 ½Iω2 = 0.145 kg × 9.80 m/s2 × 3.0 m – ½ × 0.145 kg × (6.5 m/s)2 ½Iω2 = 1.2 J