/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 144 A stone tied to the end of a str... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 144

A stone tied to the end of a string \(80 \mathrm{~cm}\) long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in \(25 \mathrm{~s}\), what is the magnitude of acceleration of the stone? (A) \(9.91 \mathrm{~m} / \mathrm{s}^{2}\) (B) \(19.82 \mathrm{~m} / \mathrm{s}^{2}\) (C) \(31 \mathrm{~m} / \mathrm{s}^{2}\) (D) None

Short Answer

Expert verified
The magnitude of acceleration of the stone is approximately \(9.91 \mathrm{~m/s}^{2}\).

Step by step solution

01

Determine the radius of the circular path

Given that the length of the string is 80 cm, this will be the radius of the circular path. We need to convert this to meters: \(r = 80 \mathrm{~cm} * \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.8 \mathrm{~m}\)
02

Calculate the number of revolutions per second

It is given that the stone makes 14 revolutions in 25 seconds. We need to find the number of revolutions per second (n): \(n = \frac{14 \mathrm{~revolutions}}{25 \mathrm{~s}} = 0.56 \mathrm{~revolutions/s}\)
03

Find the linear speed of the stone

We can now find the linear speed (v) of the stone using the formula: \(v = 2\pi rn\) Substitute the values of r and n: \(v = 2\pi(0.8\mathrm{~m})(0.56 \mathrm{~revolutions/s})\) Calculate the linear speed: \(v \approx 2.81 \mathrm{~m/s}\)
04

Calculate the centripetal acceleration

Finally, we can calculate the centripetal acceleration (a) using the formula: \(a = \frac{v^{2}}{r}\) Substitute the values of v and r: \(a = \frac{(2.81 \mathrm{~m/s})^{2}}{0.8 \mathrm{~m}}\) Calculate the centripetal acceleration: \(a \approx 9.91 \mathrm{~m/s}^{2}\) Using the steps above, we calculated the centripetal acceleration of the stone and found it to be 9.91 m/s虏, which matches option (A). Therefore, the correct answer is: (A) 9.91 m/s虏

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Most popular questions from this chapter

A particle is thrown with a speed of \(12 \mathrm{~m} / \mathrm{s}\) at an angle \(60^{\circ}\) with the horizontal range. The time interval between the moments when its speed is \(10 \mathrm{~m} / \mathrm{s}\) is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(1.0 \mathrm{~s}\) (B) \(1.2 \mathrm{~s}\) (C) \(1.4 \mathrm{~s}\) (D) \(1.6 \mathrm{~s}\)

Two particles are projected from the same point with same speed \(u\) at angles of projection \(\alpha\) and \(\beta\) from horizontal strike the horizontal ground. The maximum heights attained by projectiles is \(h_{1}\) and \(h_{2}\) respectively, \(R\) is the range for both and \(t_{1}\) and \(t_{2}\) are their time of flights respectively, then: (A) \(\alpha+\beta=\frac{\pi}{2}\) (B) \(R=4 \sqrt{h_{1} h_{2}}\) (C) \(\frac{t_{1}}{t_{2}}=\tan \alpha\) (D) \(\tan \alpha=\sqrt{h_{1} / h_{2}}\)

A particle is moving eastwards with a velocity of \(4 \mathrm{~m} / \mathrm{s}\). In \(5 \mathrm{~s}\) the velocity changes to \(3 \mathrm{~m} / \mathrm{s}\) northwards. The average acceleration in this time interval is (A) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (B) \(1 \mathrm{~m} / \mathrm{s}^{2}\) towards north-west (C) \(\frac{1}{\sqrt{2}} \mathrm{~m} / \mathrm{s}^{2}\) towards north-east (D) \(\frac{1}{2} \mathrm{~m} / \mathrm{s}^{2}\) towards north-west

A passenger reaches the platform and finds that the second last boggy of the train is passing him. The second last boggy takes \(3 \mathrm{~s}\) to pass the passenger, and the last boggy takes \(2 \mathrm{~s}\) to pass him. If passenger is late by \(t\) second from the departure of the train, then the find the value of \(2 \mathrm{t}\). (Assume that the train accelerates at constant rate and all the boggies are of equal length.)

A particle moves in a straight line with an acceleration \(\left(12 s^{2}\right) \mathrm{ms}^{-2}\) where \(s\) is the displacement of the particle in metre from \(O\), a fixed point on the line, at time \(t\) seconds. The particle has zero velocity when its displacement from \(O\) is \(-2 \mathrm{~m}\). Find the velocity (in \(\mathrm{m} / \mathrm{s}\) ) of the particle as it passes through \(O .\)
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