06 homework RAHMAN, TARIQUE Due: Feb 13 2008, 3:00 am Question 1, chap 4, sect 5. part 1 of 1 10 points Consider the following situations. A) An object moves with uniform circular motion. B) An object travels as a projectile in a gravitational eld with negligible air resistance. C) An object moves in a straight line at constant speed. In which of the situations would the object be accelerated? 1. C only 2. A only 3. A and C only 4. B and C only 5. None exhibits acceleration. 6. B only 7. A and B only correct 8. All exhibit acceleration. Explanation: A) The direction of the velocity constantly changes; the centripetal acceleration is directed toward the center of the motion. B) The projectile undergoes gratitational acceleration. C) The velocity of the object (its direction and magnitude) is unchanged, so it is not accelerated. Question 2, chap 4, sect 5. part 1 of 1 10 points A satellite of mass M moves in a circular orbit of radius R with constant speed v . True statements about this satellite include which of the following? v I) Its angular speed is . R II) Its tangential acceleration is zero. Since ar = or v= ar d 2
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III) The magnitude of its centripetal acceleration is constant. 1. II only 2. I, II, and III correct 3. I and III only 4. II and III only 5. I only Explanation: Here we will only list some facts of the circular orbital movement: v 1. The angular speed is = . R 2. The acceleration points to the center of v2 the circular orbit with magnitude = 2 R. R 3. The centripetal force is therefore the mass M times the acceleration. 2 R 2 4. The period is = . v Question 3, chap 4, sect 5. part 1 of 1 10 points A space station in the form of a large wheel, 227 m in diameter, rotates to provide an artif icial gravity of 6.7 m/s2 for people located at the outer rim. What is the frequency of the rotational motion for the wheel to produce this eect? Correct answer: 2.32012 rev/min (tolerance 1 %). Explanation: d = vt And the frequency (where T is the period) is f= 1 v = . T d 2 v2 v2 = r d
homework 06 RAHMAN, TARIQUE Due: Feb 13 2008, 3:00 am Therefore, we have ar 2 2 d (6.7 m/s2 ) 60 sec 2 2 (227 m) 1 min v
2
f= =
= 2.32012 rev/min .
At the position shown in the gure, which of the labeled arrows best represents the direction of the acceleration of the mass? 1.
Question 4, chap 4, sect 5. part 1 of 1 10 points A particle moves in the -plane xy with coordinates given by x = A cos t and y = A sin t ,
2. 3. 4. The mass is traveling at a constant velocity, therefore it has no acceleration. 5. 6.
where A = 1.5 meters and = 2.0 radians per second. What is the magnitude of the particles acceleration? 1. a = 0 m/s2 2. a = 3.0 m/s2 3. a = 4.5 m/s2
7.
8. 4. a = 1.3 m/s2 5. a = 6.0 m/s2 correct Explanation: From the expressions for x and y , it is easy to see that the particle moves on a circular orbit with a constant angular velocity and a radius of A. Therefore, the acceleration is a = 2 A = 6 m/s2 . Question 5, chap 4, sect 5. part 1 of 1 10 points A mass slides with negligible friction on a circular track of 1 m radius oriented vertically. Its speed at the position shown in the gure is 3.13 m/s. The acceleration of gravity is 9.8 m/s2 . 9. correct Explanation: The magnitude of the centripetal is v2 r (3.13 m/s)2 = 1m 9.8 m/s2 .
ar =
The centripetal acceleration is inward r . and gravity is down k
homework 06 RAHMAN, TARIQUE Due: Feb 13 2008, 3:00 am ar a g
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If the maximum value of T is 95.7 N, then we have (3.47 kg) v 2 95.7 N = . (0.465 m) From which,
Question 6, chap 4, sect 5. part 1 of 3 10 points An athlete swings a 3.47 kg ball horizontally on the end of a rope. The ball moves in a circle of radius 0.465 m at an angular speed of 0.6 rev/s. What is the tangential speed of the ball? Correct answer: 1.75301 m/s (tolerance 1 %). Explanation: = 0.6 rev/s = 3.76991 rad/s . vt = r = (0.465 m) (3.76991 rad/s) = 1.75301 m/s .
vmax = 3.58111 m/s . Question 9, chap 4, sect 5. part 1 of 1 10 points A turntable is designed to acquire an angular velocity of 44.5 rev/s in 0.7 s , starting from rest. Find the average angular acceleration of the turntable during the 0.7 s period. Correct answer: 399.431 rad/s2 (tolerance 1 %). Explanation: = 44.5 rev/s = 279.602 rad/s We have: t 279.602 rad/s = 0. 7 s = 399.431 rad/s2 .
Question 7, chap 4, sect 5. part 2 of 3 10 points What is its centripetal acceleration? Correct answer: 6.6087 m/s2 (tolerance 1 %). Explanation: ar = r 2 = (0.465 m)(3.76991 rad/s)2 = 6.6087 m/s2 .
=
Question 8, chap 4, sect 5. part 3 of 3 10 points If the maximum tension the rope can withstand before breaking is 95.7 N, what is the maximum tangential speed the ball can have? Correct answer: 3.58111 m/s (tolerance 1 %). Explanation: m v2 . F= r