Download Physics Problems & Solutions: Angular Speed, Drill, Tire Revolution, Torque, Area and more Slides Accelerator Physics in PDF only on Docsity!
Chapter 10
- (a) Find the angular speed of the Earth’s rotation about its axis. (b) How does this rotation affect the shape of the Earth? Ans: (a) The Earth rotates 2π radians (360°) on its axis in 1 day. Thus, ^ t (^2) 1 day^ rad ^ 8.641 day 10 4 s 7.27 10 ^5 rad/s (b) Because of its angular speed, the Earth bulges at the equator.
- A dentist’s drill starts from rest. After 3.20 s of constant angular acceleration, it turns at a rate of 2.51 × 104 rev/min. (a) Find the drill’s angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period. Ans: We are given ω f = 2.51 × 10^4 rev/min = 2.63 × 10 3 rad/s (a) ^^ f^ t ^ i 2.63^ ^10 3.20 s^3 rad/s^ ^0 8.21 102 rad/s^2 (b) (^) f i (^) t 12 t^2 0 12 (8.21 10 2 rad/s )(3.20 s)^2 2 4.21 103 rad
- A car accelerates uniformly from rest and reaches a speed of 22.0 m/s in 9.00 s. Assuming the diameter of a tire is 58.0 cm, (a) find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. (b) What is the final angular speed of a tire in revolutions per second? Ans: (a) We first determine the distance travelled by the car during the 9.00-s interval: s t ^ i^^ 2 ft (11.0 m/s) (9.00 s) 99.0 m the number of revolutions completed by the tire is then ^ sr (^) 0.290 m99.0 m^ 341 rad 54.3 rev (b) (^) f r^ f 22.0 m/s0.290 m^ 75.9 rad/s 12.1 rev/s
- Find the net torque on the wheel in Figure P10.27 about the axle through O , taking a = 10.0 cm and b = 25.0 cm. Ans: To find the net torque, we add the individual torques, remembering to apply the convention that a torque producing clockwise rotation is negative and a counterclockwise rotation is positive. (0.100 m)(120 N) (0.250 m)(9.00 N)(0.250 m)(10.0 N) 3.35 N m
The thirty-degree angle is unnecessary information. ANS. FIG. P10.
Chapter 11
- The displacement vectors 42.0 cm at 15.0° and 23.0 cm at 65.0° both start from the origin and form two sides of a parallelogram. Both angles are measured counterclockwise from the x axis. (a) Find the area of the parallelogram. (b) Find the length of its longer diagonal. Ans: (a)
2
area sin ( 42.0 cm )(23.0 cm sin 65. ) ( 0 15. 7
40 cm
AB
A^ ^ B
(b) The longer diagonal is equal to the sum of the two vectors.
42.0 cm cos15.0 23.0 cm cos 65.0^ ˆ 42.0 cm sin15.0 23.0 cm sin 65.0 ]ˆ
A B i j
2 2 50.3 cm ˆ^ 31.7 cmˆ length 50.3 cm 31.7 cm 59.
( ) ( ) 5 cm
A B i j A B
- The position vector of a particle of mass 2.00 kg as a function of time is given by
r^ ^ 6.00ˆ i^ 5.00 t ˆ j , where r^ ^ is in meters and t is in seconds. Determine the
angular momentum of the particle about the origin as a function of time. Ans: Differentiating r^ ^ (6.00 ˆ i^ 5.00 t ˆ j m) with respect to time gives v ^ ddt^ r^ 5.00 m/sˆ j
so p^ ^ m v 2.00 kg 5.00 m/sˆ j^ 10.0 kg m/sˆ j
and 2
6.00 5.00 0 60.0kg m / s ˆ 0 10.0 0 t i j k L^ ^ r ^ p = k
- A uniform solid disk of mass m = 3.00 kg and radius r = 0.200 m rotates about a fixed axis perpendicular to its face with angular frequency 6.00 rad/s. Calculate the magnitude of the angular momentum of the disk when the axis of rotation (a) passes through its center of mass and (b) passes through a point midway between the center and the rim. Ans: (a) For an axis of rotation passing through the center of mass, the magnitude of the angular momentum is given by (^2) ^2 2
(^12 1 2) 3.00 kg 0.200 m 6.00 rad / s 0.360 kg m / s
L I ^ MR
(b) For a point midway between the center and the rim, we use the parallel-axis theorem to find the moment of inertia about this point. Then,
2 2 (^2 )
(^3 4) 3.00 kg 0.200 m 6.00 rad / s 0.540 kg m / s
L I ^ MR M ^^ R
^
- A playground merry-go-round of radius R = 2.00 m has a moment of inertia I = 250 kg · m^2 and is rotating at 10.0 rev/min about a frictionless, vertical axle. Facing the axle, a 25.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round? Ans: From conservation of angular momentum, I (^) i ωi = Ifωf : (250 kg · m^2 )(10.0 rev/min) = [250 kg · m^2 + (25.0 kg)(2.00 m) 2 ] ω 2 ω 2 = 7.14 rev/min
