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Chapter 3: Problem 78

For a particle rotating in a vertical circle with uniform speed, the maximum and minimum tension in the string are in the ratio \(5: 3\). If the radius of vertical circle is \(2 \mathrm{~m}\), the speed of revolving body is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(\sqrt{5} \mathrm{~m} / \mathrm{s}\) (B) \(4 \sqrt{5} \mathrm{~m} / \mathrm{s}\) (C) \(5 \mathrm{~m} / \mathrm{s}\) (D) \(10 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The speed of the revolving body is \(4\sqrt{5}\) m/s.

Step by step solution

01

Tension at the top of the circle

At the top of the vertical circle, the centripetal force is the difference between the tension and the gravitational force acting downwards. We can represent this as: \(T_{min} = mv^2/R - mg \), where \(T_{min}\) is the minimum tension, \(m\) is the mass of the particle, \(v\) is its speed, \(R\) is the radius and, \(g\) is the gravitational acceleration.
02

Tension at the bottom of the circle

At the bottom of the circle, the tension is the sum of the centripetal force and the gravitational force acting downwards. We can represent this as: \(T_{max} = mv^2/R + mg \), where \(T_{max}\) represents the maximum tension.
03

Using the given tension ratio

We're given that the ratio of maximum to minimum tension is 5:3. We can now use this information to form an equation: \(\frac{T_{max}}{T_{min}} = \frac{5}{3}\). Substituting the previous equations for the tensions, we get: \(\frac{mv^2/R + mg}{mv^2/R - mg} = \frac{5}{3}\).
04

Simplify the equation and solve for v

Let's now simplify the equation and solve it for the speed v: \(\frac{mv^2 + mRg}{mv^2 - mRg} = \frac{5}{3}\) Cross-multiplying, we get: \(3mv^2 + 3mRg = 5mv^2 - 5mRg\) Bringing all the terms to one side, we have: \(2mv^2 = 8mRg\) Now, divide both sides by \(2m\) to isolate the speed v: \(v^2 = 4Rg\) Finally, take the square root of both sides: \(v = 2\sqrt{Rg}\)
05

Substitute the values and find the speed

Now that we have the expression for the speed of the revolving body, substitute the given radius (R = 2 m) and gravitational acceleration (g = 10 m/s虏) into the formula: \(v = 2\sqrt{2 * 10}\) \(v = 2\sqrt{20}\) \(v = 4\sqrt{5}\) Hence, the speed of the revolving body is \(4\sqrt{5}\) m/s, which corresponds to option (B).

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Most popular questions from this chapter

Two masses \(8 \mathrm{~kg}\) and \(12 \mathrm{~kg}\) are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the tension the string when the masses are released. (A) \(96 \mathrm{~N}\) (B) \(80 \mathrm{~N}\) (C) \(100 \mathrm{~N}\) (D) None of these

A girl of mass \(50 \mathrm{~kg}\) stands on a measuring scale in a lift. At an instant, it is detected that the reading reduces to \(40 \mathrm{~kg}\) for a while and then returns to original value. It can be said that (A) The lift was in constant motion upwards (B) The lift was in constant motion downwards (C) The lift was suddenly started in downward motion (D) The lift was suddenly started in upward motion

Two blocks of masses \(5 \mathrm{~kg}\) and \(3 \mathrm{~kg}\) are placed on a smooth horizontal surface. A horizontal force \(F=16 \mathrm{~N}\) is applied on \(5 \mathrm{~kg}\) as shown. Find normal force between the blocks.

A particle is moving with a constant angular acceleration of \(4 \mathrm{rad} / \mathrm{s}^{2}\) in a circular path. At \(t=0\), particle was at rest. Find the time at which the magnitudes of centripetal acceleration and tangential acceleration are equal. \(\begin{array}{llll}\text { (A) } 1 \mathrm{~s} & \text { (B) } 2 \mathrm{~s} & \text { (C) } \frac{1}{2} \mathrm{~s} & \text { (D) } \frac{1}{4} \mathrm{~s}\end{array}\)

Four forces act on a point object. The object will be in equilibrium if (A) all of them are in the same plane (B) they are opposite to each other in pair (C) the sum of \(x, y\) and \(z\) components of all the force is zero separately (D) they are from a closed figure of four sides
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