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Chapter 6: Problem 80

The angular momentum of a particle rotating with a central force is constant due to (A) constant linear momentum. (B) zero torque. (C) constant torque. (D) constant force.

Short Answer

Expert verified
(B) Zero Torque is the reason for the constant angular momentum in this case.

Step by step solution

01

Analyzing Options

First, go through each of the given options: (A) constant linear momentum, (B) zero torque, (C) constant torque, and (D) constant force, and analyze their meaning and implications.
02

Rule out Irrelevant Options

Next, rule out options that do not comply with the concept of the constant angular momentum of a particle. In this case, Angular momentum is not related directly to constant linear momentum (A) and constant force (D), so these options can be ruled out.
03

Choose Relevant Option

Between zero torque (B) and constant torque (C), choosing the correct option is critical. Torque tends to change angular momentum. A constant torque would not maintain a constant angular momentum because it would cause a continuous change in angular momentum. The torque has to be zero in order to conserve the angular momentum.

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Most popular questions from this chapter

An annular ring with inner and outer radii \(R_{1}\) and \(R_{2}\) is rolling without slipping with a uniform angular speed. The ratio of the forces experienced by two particles situated on the inner and outer parts of the ring is (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) \([2005]\) (A) \(\frac{R_{1}}{R_{2}}\) (B) 1 (D) \(\frac{R_{2}}{R_{1}}\) [Note: The particles should be of same mass]

Two balls of mass \(M=9 \mathrm{~g}\) and \(m=3 \mathrm{~g}\) are attached by massless threads \(A O\) and \(O B\). The length \(A B\) is \(1 \mathrm{~m}\). They are set in rotational motion in a horizontal plane about a vertical axis at \(O\) with constant angular velocity \(\omega\). The ratio of length \(A O\) and \(O B\left(\frac{O B}{A O}\right)\) for which the tension in threads are same will be.

A hoop of radius \(R\) and mass \(m\) rotating with an angular velocity \(\omega_{0}\) is placed on a rough horizontal surface. The initial velocity of the centre of the hoop is zero. What will be the velocity if the centre of the loop ceases to slip? (A) \(\frac{r \omega_{0}}{3}\) (B) \(\frac{r \omega_{0}}{2}\) (C) \(r \omega_{0}\) (D) \(\frac{r \omega_{0}}{4}\)

A circular disc of radius \(R\) is removed from a bigger circular disc of radius \(2 R\) such that the circumferences of the discs coincide. The centre of mass of the new disc is \(a / R\) from the centre of the bigger disc. The value of \(a\) is (A) \(\frac{1}{2}\) (B) \(\frac{1}{6}\) (C) \(\frac{1}{4}\) (D) \(\frac{1}{3}\)

Let \(a_{r}\) and \(a_{t}\) represent radial and tangential acceleration. The motion of a particle may be circular if (A) \(a_{r}=a_{t}=0\) (B) \(a_{r}=0\) and \(a_{t} \neq 0\) (C) \(a_{r} \neq 0\) and \(a_{t}=0\) (D) \(a_{r} \neq 0\) and \(a_{t} \neq 0\)
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