Angular Velocity and Final Tuesday
Q1. Figure 1 shows a solid cylindrical steel rod of length = 2.0 m and diameter D = 2.0 cm. What will be increase in its length when m = 80 kg block is attached to its bottom end? (Young's modulus of steel = 1.9 × 1011 Pa) Fig#
Answer:
∆L = FL mg = AY AY =
= 0.0000262689 m
A) B) C) D) E)
2.6 x 10-5 m 1.3 x 10-5 m 4.8 x 10-5 m 7.2 x 10-5 m 3.5 x 10-5 m
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 1
Q2. In Fig. 2, PQ is a horizontal uniform beam weighing 155 N. It is supported by a string and a hinge at point P. A 245 N block is hanging from point Q at the end of the beam. Find the horizontal component of net force on the beam from the hinge.
Answer:
Toque about P implies:
0 (T sin 35 ) (1.35) − (155) 1.7 − (245)(1.7) = 2 o = ⇒T
131.75 + 416.5 = 708 N (1.35)sin 35o
cos35o cos35o ) = T= 708 (= 579.98 N ≈ 580 N FH
The force on x-axis implies:
A) B) C) D) E)
580 N 310 N 491 N 164 N 200 N
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 2
Q3. A 20.0 m long uniform beam weighing 550 N rests on supports “A” and “B”, as shown in Figure 3. Find the magnitude of the force that the support “A” exerts on the beam when the block of weight 200 N is placed at D. Fig#
Answer:
The torque about point B implies:
− M D × 5 + M beam × 5 − FA × 12 = 0 = FA
A) B) C) D) E)
−200 × 5 + 550 × 5 = 145.8 N ≈ 146 N 12
146 N 241 N 501 N 315 N 185 N
Q4. At what height above earth’s surface would the gravitational acceleration be 0.980 m/s2?
Answer:
2 GM g GM / R E = = ⇒h = (R E + h ) 2 10 10
10R E − R E = 1.38 × 107
7 = 1.37737 10 m
A) B) C) D) E) 1.38 × 107 m 1.12 × 107 m 7.12 × 107 m 5.82 × 108 m 4.05 × 108 m
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 3
Q5. In Figure 4, what is the net gravitational force exerted on the 5.00 kg uniform sphere by the other two uniform spheres? Fig#
Answer:
The gravitational force is
Answer:
∆L = FL mg = AY AY =
= 0.0000262689 m
A) B) C) D) E)
2.6 x 10-5 m 1.3 x 10-5 m 4.8 x 10-5 m 7.2 x 10-5 m 3.5 x 10-5 m
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 1
Q2. In Fig. 2, PQ is a horizontal uniform beam weighing 155 N. It is supported by a string and a hinge at point P. A 245 N block is hanging from point Q at the end of the beam. Find the horizontal component of net force on the beam from the hinge.
Answer:
Toque about P implies:
0 (T sin 35 ) (1.35) − (155) 1.7 − (245)(1.7) = 2 o = ⇒T
131.75 + 416.5 = 708 N (1.35)sin 35o
cos35o cos35o ) = T= 708 (= 579.98 N ≈ 580 N FH
The force on x-axis implies:
A) B) C) D) E)
580 N 310 N 491 N 164 N 200 N
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 2
Q3. A 20.0 m long uniform beam weighing 550 N rests on supports “A” and “B”, as shown in Figure 3. Find the magnitude of the force that the support “A” exerts on the beam when the block of weight 200 N is placed at D. Fig#
Answer:
The torque about point B implies:
− M D × 5 + M beam × 5 − FA × 12 = 0 = FA
A) B) C) D) E)
−200 × 5 + 550 × 5 = 145.8 N ≈ 146 N 12
146 N 241 N 501 N 315 N 185 N
Q4. At what height above earth’s surface would the gravitational acceleration be 0.980 m/s2?
Answer:
2 GM g GM / R E = = ⇒h = (R E + h ) 2 10 10
10R E − R E = 1.38 × 107
7 = 1.37737 10 m
A) B) C) D) E) 1.38 × 107 m 1.12 × 107 m 7.12 × 107 m 5.82 × 108 m 4.05 × 108 m
Phys101 Term: 111
Final Tuesday, January 10, 2012
Code: 1 Page: 3
Q5. In Figure 4, what is the net gravitational force exerted on the 5.00 kg uniform sphere by the other two uniform spheres? Fig#
Answer:
The gravitational force is