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Chapter 9: Problem 7

A particle moves in a circle uniformly such that it completes a circle in \(2 \mathrm{sec}\). During half a turn its centripetal acceleration change by \(6 \mathrm{~m} / \mathrm{s}^{2}\). Choose the correct option(s). (1) the radius of the circle is \(\left(3 / \pi^{2}\right) \mathrm{m}\). (2) the magnitude of change in velocity during \(1 \mathrm{~s}\) is \(3 / \pi \mathrm{m} / \mathrm{s}\) (3) the tangential velocity of the particle is \(3 / \pi \mathrm{m} / \mathrm{s}\). (4) the particle is moving in anticlockwise sense as seen from above.

Short Answer

Expert verified
Options (1), (2), and (3) are correct.

Step by step solution

01

Calculate the Tangential Velocity

The particle completes a circle in 2 seconds, which means its period, \( T \), is 2 seconds. The tangential velocity, \( v \), can be calculated using the formula for the circumference of a circle \( v = \frac{2\pi R}{T} \). We need to find the expression for velocity using the given centripetal acceleration change.
02

Determine Centripetal Acceleration

The centripetal acceleration, \( a_c \), is given by \( v^2/R \). Given that this acceleration changes by \( 6 \, \text{m/s}^2 \) during half a turn, we have that the change in centripetal acceleration is due to change from \( 0 \) to maximum, hinting \[ 2 \cdot \frac{v^2}{R} = 6 \Rightarrow v^2 = 3R \].
03

Solve for the Radius

Using the equations \( v = \frac{2\pi R}{2} = \pi R \) and \( v^2 = 3R \), set \( v = \pi R \) into \( v^2 = 3R \): \[ (\pi R)^2 = 3R \] \[ \pi^2 R^2 = 3R \] \[ R = \frac{3}{\pi^2} \]
04

Calculate Magnitude of Change in Velocity

The change in velocity during half the circle (1 second) is \[ \Delta v = v - (-v) = 2v \]. From \( v = \pi R \), substituting the radius \( R = \frac{3}{\pi^2} \): \[ v = 3/\pi \] and hence \( \, \Delta v = 2(3/\pi) = 6/\pi \). Thus, in 1 second (half rotation), \( \, \Delta v = 3/\pi \) m/s.
05

Confirm Options

Option (1) is correct as \( R = \frac{3}{\pi^2} \) m. Option (2) is true since change in velocity in 1 second is \( 3/\pi \). Option (3) is correct as the tangential velocity is also \( 3/\pi \). Option (4) cannot be confirmed with this information only.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Acceleration
When an object moves in a circular path at constant speed, it experiences centripetal acceleration. This is the acceleration that keeps the object moving in a circle, directing it towards the center of the circle. The magnitude of this acceleration is given by the formula \( a_c = \frac{v^2}{R} \), where \( v \) is the tangential velocity and \( R \) is the radius of the circular path.

In the context of this exercise, the centripetal acceleration of the particle changes due to its completion of half a circle. Essentially, this means the centripetal acceleration doubles as the particle transitions from being at rest to its maximum acceleration. This results in a change of 6 m/s虏, giving us the equation \( 2 \cdot \frac{v^2}{R} = 6 \) which helps us solve for the radius and velocity.
Tangential Velocity
Tangential velocity refers to the linear speed of an object moving along a circular path. It is directed along the tangent to the path at any point and is a crucial component in calculating dynamic properties of circular motion.

The formula to determine tangential velocity is \( v = \frac{2\pi R}{T} \), where \( T \) is the period, or time taken to complete one full circle. In this exercise, since the particle finishes a complete circle in 2 seconds, the formula simplifies using the derived value of the radius, resulting in \( v = \pi R \). With the radius calculated as \( \frac{3}{\pi^2} \), the tangential velocity equates to \( \frac{3}{\pi} \) m/s. Hence, options regarding the tangential velocity in this setup are verified as true.
Radius of Circular Path
The radius of a circular path is a critical element in uniform circular motion, impacting both centripetal acceleration and tangential velocity. In our given exercise, we determine the radius by using the expression derived from changes in acceleration. By equating \( v^2 = 3R \) and combining it with \( v = \pi R \), the expression for the radius is solved as \( R = \frac{3}{\pi^2} \) meters.

This calculated radius helps not only in verifying the correct options in the exercise but also provides insight into the relationship between linear speed and radial distance in circular motion. A smaller radius typically indicates a higher centripetal force requirement to maintain circular motion, illustrating how closely tied these properties are.

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Most popular questions from this chapter

A body is moving in a circle with a speed of \(1 \mathrm{~ms}^{-1}\). This speed increases at a constant rate of \(2 \mathrm{~ms}^{-1}\) every second. Assume that the radius of the circle described is \(25 \mathrm{~m}\). The total acceleration of the body after \(2 \mathrm{~s}\) is (1) \(2 \mathrm{~ms}^{-2}\) (2) \(25 \mathrm{~ms}^{-2}\) (3) \(\sqrt{5} \mathrm{~m} \mathrm{~s}^{-2}\) (4) \(\sqrt{7} \mathrm{~m} \mathrm{~s}^{-2}\)

The track of motorcycle-race is circular and unbanked. There are two bikers on the road, one travels along a path of greater radius than the other. They both lean towards the centre at the same angle. Which one completes the circular path in less time? (1) biker having path of smaller radius. (2) biker having path of larger radius. (3) will complete circle in same time. (4) both can not lean at same angle if radius is different.

A small block of mass \(m\) slides on a frictionless horizontal table. It is constrained to move inside a ring of radius \(l\) which is fixed to the table. At \(t=0\), the block has a tangential velocity \(v_{0}\). The coefficient of friction between the block and the ring is \(\mu\). The velocity of the block at time \(t\) is (I) \(\frac{v_{0}}{1+\frac{\mu v_{0} t}{l}}\) (2) \(\frac{v_{0}}{1+\frac{2 \mu v_{0} t}{l}}\) (3) \(\frac{v_{0}}{1-\left(\frac{\mu v_{0} t}{l}\right)}\) (4) \(\frac{v_{0}}{\left(\frac{\mu v_{0} t}{l}\right)}\)

A small ring \(P\) is threaded on a smooth wire bent in the form of a circle of radius \(a\) and center \(O .\) The wire is rotating with constant angular speed \(\omega\) about a vertical diameter \(X Y\), while the ring remains at rest relative to the wire at a distance \(a / 2\) from \(X Y\). Then \(\omega^{2}\) is equal to (1) \(\frac{2 g}{a}\) (2) \(\frac{g}{2 a}\) (3) \(\frac{2 g}{a \sqrt{3}}\) (4) \(\frac{g \sqrt{3}}{2 a}\)

A particle is moving along a circular path. The angul:d velocity, linear velocity, angular acceleration, and centripetal acceleration of the particle at any instant, respectively, ari \(\vec{\omega}, \vec{v}, \vec{\alpha}\), and \(\vec{a}_{c^{*}}\) Which of the following relations is noid correct? (1) \(\vec{\omega} \perp \vec{v}\) (2) \(\vec{\omega} \perp \vec{\alpha}\) (3) \(\vec{\omega} \perp \vec{\alpha}_{c}\) (4) \(\vec{v} \perp \vec{\alpha}_{c}\)
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