/*! 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 63 One end of a string of length \(... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 63

One end of a string of length \(l\) is connected to a particle of mass \(m\) and the other to a small peg on a smooth horizontal table. If the particle moves in a circular motion with speed \(v\), the net force on the particle (directed towards the centre) is: (A) \(T\) (B) \(T-\frac{m v^{2}}{l}\) (C) \(T+\frac{m v^{2}}{l}\) (D) 0

Short Answer

Expert verified
(A) \(T\)

Step by step solution

01

Write down the given information and the formula for centripetal force

We are given: - Mass of the particle, \(m\) - Length of the string, \(l\) - Speed of the particle, \(v\) The formula for centripetal force (\(F_c\)) acting on a particle moving in a circle of radius \(r\) with speed \(v\) is given by: \(F_c = \frac{m v^2}{r}\) In this case, the radius of the circle is equal to the length of the string, \(l\). So, the centripetal force (directed towards the center) acting on the particle is: \(F_c = \frac{m v^2}{l}\)
02

Compare the centripetal force with the given options

Now let's compare the centripetal force, \(\frac{m v^2}{l}\), with the given options: (A) \(T\) (B) \(T - \frac{m v^2}{l}\) (C) \(T + \frac{m v^2}{l}\) (D) 0 We can see that none of the options match the given centripetal force directly. However, considering that the net force directed towards the center would be equal to the centripetal force, we can safely assume that the tension in the string (\(T\)) must be equal to the centripetal force acting on the particle. In other words, the tension in the string provides the centripetal force required to keep the particle moving in a circle. Therefore, the net force acting on the particle towards the center is equal to the centripetal force, which is equal to the tension in the string (\(T\)). The correct answer is: (A) \(T\).

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Most popular questions from this chapter

A block of mass \(10 \mathrm{~kg}\) is placed on a rough inclined plane of inclination \(37^{\circ}\left(\tan 37^{\circ}=3 / 4\right)\). The co-efficient of friction between block and surface is \(0.4\). A horizontal force \(F=50 \mathrm{~N}\) is applied on the block, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block is zero. (B) Acceleration of block is \(2.4 \mathrm{~m} / \mathrm{s}^{2}\) along the inclined plane. (C) Frictional force between block and surface is \(44 \mathrm{~N}\). (D) Frictional force between block and surface is \(20 \mathrm{~N}\).

The upper half of an incline plane with inclination \(\phi\) is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the co-efficient of friction for the lower half is given by (A) \(2 \tan \phi\) (B) \(\tan \phi\) (C) \(2 \sin \phi\) (D) \(2 \cos \phi\)

An elevator starts from rest with a constant upward acceleration. It moves \(2 \mathrm{~m}\) in the first \(0.6\) second. \(\mathrm{A}\) passenger in the elevator is holding a \(3 \mathrm{~kg}\) package by a vertical string. When the elevator is moving, what is the tension in the string? (A) \(4 \mathrm{~N}\) (B) \(62.7 \mathrm{~N}\) (C) \(29.4 \mathrm{~N}\) (D) \(20.6 \mathrm{~N}\)

A metre scale is moving with uniform velocity. This implies (A) the force acting on the scale is zero, but a torque about the centre of mass can act on the scale. (B) the force acting on the scale is zero and the torque acting about centre of mass of the scale is also zero. (C) the total force acting on it need not be zero but the torque on it is zero. (D) neither the force nor the torque needs to be zero.

A block \(P\) of mass \(4 \mathrm{~kg}\) is placed on horizontal rough surface with co-efficient of friction \(\mu=0.6\). And two blocks \(R\) and \(Q\) of masses \(2 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) connected with the help of massless strings \(A\) and \(B\), respectively, passing over frictionless pulleys as shown, then \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) Acceleration of block \(P\) is zero. (B) Tension in string \(A\) is \(20 \mathrm{~N}\). (C) Tension in string \(B\) is \(40 \mathrm{~N}\). (D) Contact force on block \(P\) is \(20 \sqrt{5} \mathrm{~N}\).
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