University:
Georgia Southern University
Course:
MATH 2242 | Calculus II
Academic year:

2024
Views:

32

Pages:

4
Author:

Jarrett Brock

Question 1 A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.200 kg 0.200kg, and its center of gravity is located at its geometrical center. On the tray is a 1.00 kg 1.00kg plate of food and a 0.190 kg 0.190kg cup of coffee. Obtain the force 𝑇 T exerted by the thumb and the force 𝐹 F exerted by the four fingers. Both forces act perpendicular to the tray, which is being held parallel to the ground. Answer: Since the entire system is in equilibrium, the net torque about any point due to all forces acting on it must be zero. Taking the net torque about the four fingers (where 𝐹 F is acting upwards): ( 𝑇 × 0.04 m ) − ( 1 kg × 0.14 m × 𝑔 ) − ( 0.19 kg × 0.32 m × 𝑔 ) = 0 , (T×0.04m)−(1kg×0.14m×g)−(0.19kg×0.32m×g)=0, with 𝑔 = 9.8 m/s 2 g=9.8m/s 2 . Solving: 𝑇 = 5.22 N upwards . T=5.22N upwards. Similarly, net torque about the thumb: ( 𝐹 × 0.04 m ) − ( 0.2 kg × 0.14 m × 𝑔 ) − ( 1 kg × 0.18 m × 𝑔 ) − ( 0.19 kg × 0.32 m × 𝑔 ) = 0. (F×0.04m)−(0.2kg×0.14m×g)−(1kg×0.18m×g)−(0.19kg×0.32m×g)=0. Solving: 𝐹 = 6.5 N upwards . F=6.5N upwards. Question 2 The rotor (flywheel) of a toy gyroscope has a mass of 0.140 kg 0.140kg, a moment of inertia about its axis 𝐼 I, and a mass of the frame of 0.0250 kg 0.0250kg. The gyroscope is supported on a single pivot with its center of mass at a horizontal distance of 0.040 m 0.040m from the pivot. The gyroscope is precessing in the horizontal plane at the rate of one revolution in 2.20 s 2.20s. Find the angular speed at which the rotor spins about its axis, expressed in rev/min rev/min. Answer: Force exerted by the pivot is: 𝑃 = 𝑀 × 𝑔 × 𝑟 𝜔 . P= ω M×g×r . 𝑀 M is the mass of the flywheel = 0.140 kg =0.140kg, 𝑟 = 0.04 m r=0.04m, and 𝜔 = 2 𝜋 / 𝑇 = 2 𝜋 / 2.2 = 2.85 rad/s ω=2π/T=2π/2.2=2.85rad/s. Solving using the above values yields the angular speed of the rotor. Question 3 An exceptional standing jump would raise a person 0.80 m 0.80m off the ground. To do this, what force must a 66 kg 66kg person exert against the ground? Assume the person crouches a distance of 0.20 m 0.20m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground. Answer: Given: Jump height ℎ = 0.80 m h=0.80m, Distance of crouch 𝑑 = 0.20 m d=0.20m, Mass 𝑚 = 66 kg m=66kg. Initial velocity is 𝑢 = 0 u=0. Use the second equation of motion: ℎ = 𝑣 2 2 𝑔 , 𝑣 = 2 𝑔 ℎ = 2 ( 9.81 ) ( 0.80 ) = 3.96 m/s . h= 2g v 2 ,v= 2gh = 2(9.81)(0.80) =3.96m/s. Use 𝐹 = 𝑚 𝑎 F=ma: 𝑎 = 𝑣 2 2 𝑑 = ( 3.96 ) 2 2 ( 0.20 ) = 39.24 m/s 2 . a= 2d v 2 = 2(0.20) (3.96) 2 =39.24m/s 2 . 𝐹 = 𝑚 × 𝑎 = 66 × 39.24 = 2587.5 N . F=m×a=66×39.24=2587.5N. Question 4 A car is initially going 50 ft/s 50ft/s and brakes at a constant rate (constant negative acceleration), coming to a stop in 5 s 5s. (a) Graph the velocity from 𝑡 = 0 t=0 to 𝑡 = 5 t=5. Answer: Use the equation: 𝑣 = 𝑢 + 𝑎 𝑡 , v=u+at, where 𝑢 = 50 ft/s , 𝑣 = 0 , 𝑡 = 5 s u=50ft/s,v=0,t=5s. Solve for 𝑎 a: 0 = 50 + 𝑎 ( 5 ) ⟹ 𝑎 = − 10 ft/s 2 . 0=50+a(5)⟹a=−10ft/s 2 . Plot 𝑣 = 50 − 10 𝑡 v=50−10t from 𝑡 = 0 t=0 to 𝑡 = 5 t=5. At 𝑡 = 0 , 𝑣 = 50 t=0,v=50; At 𝑡 = 5 , 𝑣 = 0 t=5,v=0. The graph is a straight line with a negative slope from ( 0 , 50 ) (0,50) to ( 5 , 0 ) (5,0).