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Chapter 1 SP 014 Exercise 1 1. A particle is moving on a circular path of radi...
Aug 30, 2024
Solution
a
To find the time taken to complete 20 revolutions, we first calculate the time for one revolution. The speed of the particle is given as 10 ms−110 \mathrm{~ms}^{-1} and the radius is 0.5 m0.5 \mathrm{~m}. The circumference of the circular path is given by C=2πr=2π(0.5)=π m.C = 2\pi r = 2\pi(0.5) = \pi \mathrm{~m}. The time for one revolution is then calculated using the formula tone=Cv=π10≈0.314 s.t_{one} = \frac{C}{v} = \frac{\pi}{10} \approx 0.314 \mathrm{~s}. Therefore, the time for 20 revolutions is ttotal=20×tone=20×0.314≈6.28 s.t_{total} = 20 \times t_{one} = 20 \times 0.314 \approx 6.28 \mathrm{~s}.
b
The angular velocities of the hour-hand and minute-hand of a clock can be calculated using the formula for angular velocity ω=2πT\omega = \frac{2\pi}{T} where TT is the period. For the hour-hand, T=12 h=43200 sT = 12 \mathrm{~h} = 43200 \mathrm{~s}, thus ωhour=2π43200≈1.45×10−4rad/s.\omega_{hour} = \frac{2\pi}{43200} \approx 1.45 \times 10^{-4} \mathrm{rad/s}. For the minute-hand, T=1 h=3600 sT = 1 \mathrm{~h} = 3600 \mathrm{~s}, thus ωminute=2π3600≈1.75×10−3rad/s.\omega_{minute} = \frac{2\pi}{3600} \approx 1.75 \times 10^{-3} \mathrm{rad/s}.
c
The angular velocity can be calculated using the formula ω=vr\omega = \frac{v}{r} where v=6.0 ms−1v = 6.0 \mathrm{~ms}^{-1} and the radius r=2.52=1.25 mr = \frac{2.5}{2} = 1.25 \mathrm{~m}. Thus, ω=6.01.25=4.8rad/s.\omega = \frac{6.0}{1.25} = 4.8 \mathrm{rad/s}.
d
The number of complete revolutions per minute (rpm) can be calculated from the time taken for one revolution. The time for one revolution is given as 0.25 s0.25 \mathrm{~s}, so the number of revolutions per minute is rpm=60 s0.25 s=240 rpm.\text{rpm} = \frac{60 \mathrm{~s}}{0.25 \mathrm{~s}} = 240 \mathrm{~rpm}.
e
For a particle moving in a circle with a diameter of 5.00 cm5.00 \mathrm{~cm}, the radius is 0.025 m0.025 \mathrm{~m}. The angular velocity can be calculated using the formula ω=2πT\omega = \frac{2\pi}{T} where T=0.5 sT = 0.5 \mathrm{~s} (since it takes 3.0 s to move from one end to the other). Thus, ω=2π0.5=1.05rads−1.\omega = \frac{2\pi}{0.5} = 1.05 \mathrm{rads}^{-1}. The speed can be calculated using v=rω=0.025×1.05=0.0262 ms−1.v = r\omega = 0.025 \times 1.05 = 0.0262 \mathrm{~ms}^{-1}.
f
For an object moving in a circle of radius 0.4 m0.4 \mathrm{~m} with an angular velocity of 6rads−16 \mathrm{rads}^{-1}, the linear velocity is given by v=rω=0.4×6=2.4 ms−1.v = r\omega = 0.4 \times 6 = 2.4 \mathrm{~ms}^{-1}. The time for one complete revolution is given by T=2πω=2π6≈1.05 s.T = \frac{2\pi}{\omega} = \frac{2\pi}{6} \approx 1.05 \mathrm{~s}.
g
The angular velocity can be calculated as follows: The mass completes 10 revolutions in one second, so the angular velocity is ω=10×2π=62.8rads−1.\omega = 10 \times 2\pi = 62.8 \mathrm{rads}^{-1}. The linear velocity can be calculated using v=rω=0.8×62.8=50.3 ms−1.v = r\omega = 0.8 \times 62.8 = 50.3 \mathrm{~ms}^{-1}.
h
The number of revolutions made by a disc with a diameter of 0.3 m0.3 \mathrm{~m} rolling 65 m65 \mathrm{~m} can be calculated using the circumference C=πd=π(0.3)≈0.942 m.C = \pi d = \pi(0.3) \approx 0.942 \mathrm{~m}. The number of revolutions is revolutions=650.942≈69 rev.\text{revolutions} = \frac{65}{0.942} \approx 69 \mathrm{~rev}. The angular displacement is given by θ=revolutions×2π=69×2π≈138π rad.\theta = \text{revolutions} \times 2\pi = 69 \times 2\pi \approx 138\pi \mathrm{~rad}.
i
The angular displacement, angular speed, and angular acceleration can be determined by differentiating the angular displacement function. At t=0 st=0 \mathrm{~s}, θ(0)=5.00 rad,ω(0)=10.0 rads−1,α(0)=4.00 rads−2.\theta(0) = 5.00 \mathrm{~rad}, \quad \omega(0) = 10.0 \mathrm{~rads}^{-1}, \quad \alpha(0) = 4.00 \mathrm{~rads}^{-2}. At t=3.00 st=3.00 \mathrm{~s}, θ(3)=5.00+10.0(3)+2.00(32)=53.0 rad,ω(3)=10.0+2.00(3)=16.0 rads−1,α(3)=4.00 rads−2.\theta(3) = 5.00 + 10.0(3) + 2.00(3^2) = 53.0 \mathrm{~rad}, \quad \omega(3) = 10.0 + 2.00(3) = 16.0 \mathrm{~rads}^{-1}, \quad \alpha(3) = 4.00 \mathrm{~rads}^{-2}.
j
For a disk rotating at 1200rev/min1200 \mathrm{rev/min}, we first convert to radians per second: ω=1200×2π60=125.66 rads−1.\omega = 1200 \times \frac{2\pi}{60} = 125.66 \mathrm{~rads}^{-1}. The tangential speed at a point 3.00 cm3.00 \mathrm{~cm} from the center is given by v=rω=0.03×125.66≈3.77 ms−1.v = r\omega = 0.03 \times 125.66 \approx 3.77 \mathrm{~ms}^{-1}. The radial (centripetal) acceleration is given by ac=v2r=(3.77)20.08≈177.5 m/s2.a_c = \frac{v^2}{r} = \frac{(3.77)^2}{0.08} \approx 177.5 \mathrm{~m/s^2}. The total distance moved in 2.00 s2.00 \mathrm{~s} is d=v×t=3.77×2.00≈7.54 m.d = v \times t = 3.77 \times 2.00 \approx 7.54 \mathrm{~m}.
k
For a grinding wheel rotating at 2500 rpm2500 \mathrm{~rpm}, we convert to radians per second: ω=2500×2π60≈261.8 rads−1.\omega = 2500 \times \frac{2\pi}{60} \approx 261.8 \mathrm{~rads}^{-1}.
[1] Answer
6.28 s
[2] Answer
1.45 × 10^{-4} rad/s ; 1.75 × 10^{-3} rad/s
[3] Answer
4.8 rad/s
[4] Answer
240 rpm
[5] Answer
1.05 rads^{-1} ; 2.62 × 10^{-2} ms^{-1}
[6] Answer
2.4 ms^{-1} ; 1.05 s
[7] Answer
62.8 rads^{-1} ; 50.3 ms^{-1}
[8] Answer
69 rev ; 138 π rad
[9] Answer
5.00 rad ; 10.0 rads^{-1} ; 4.00 rads^{-2} ; 53.0 rad ; 22.0 rads^{-1} ; 4.00 rads^{-2}
[10] Answer
125.66 rads^{-1} ; 3.77 ms^{-1} ; 177.5 m/s^2 ; 7.54 m
[11] Answer
261.8 rads^{-1}
Key Concept
Circular Motion: The study of objects moving in a circular path, involving concepts such as angular velocity, tangential velocity, centripetal acceleration, and forces acting on the object. Key equations include Fc=mv2rF_c = \frac{mv^2}{r} for centripetal force and ω=vr\omega = \frac{v}{r} for angular velocity.
Explanation
The solutions provided utilize fundamental principles of circular motion to calculate various parameters such as time, angular velocity, and linear velocity based on the given conditions. Each calculation follows the relevant equations for circular motion.
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