Question:
One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.
The total kinetic energy of the ring is
One twirls a circular ring (of mass $M$ and radius $R$ ) near the tip of one's finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is $r$. The finger rotates with an angular velocity $\omega_{0}$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.
The total kinetic energy of the ring is
Updated On: Jan 1, 2023
- $M \omega_{0}^{2} R^{2}$
- $\frac{1}{2} M \omega_{0}^{2}(R-r)^{2}$
- $M \omega_{0}^{2}(R-r)^{2}$
- $\frac{3}{2} M \omega_{0}^{2}(R-r)^{2}$
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Answer (c) $M \omega_{0}^{2}(R-r)^{2}$
Was this answer helpful?
2
0
Top Questions on Gravitation
- At a height \( h \) above the Earth's surface, the acceleration due to gravity becomes \( \frac{g}{\sqrt{3}} \). What is the value of \( h \) in terms of the Earth's radius \( R \)?
- MHT CET - 2025
- Physics
- Gravitation
- Two objects of masses 5 kg and 10 kg are placed 2 meters apart. What is the gravitational force between them?
(Use \(G = 6.67 \times 10^{-11}\, \mathrm{Nm^2/kg^2}\))- AP EAPCET - 2025
- Physics
- Gravitation
- The escape velocity from the surface of a planet is \(v_e\). What will be the escape velocity from a planet whose mass and radius are twice that of the original planet?
- BITSAT - 2025
- Physics
- Gravitation
- Consider a star of mass $ m_2 $ kg revolving in a circular orbit around another star of mass $ m_1 $ kg with $ m_1 \gg m_2 $. The heavier star slowly acquires mass from the lighter star at a constant rate of $ \gamma $ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $ r $, then its relative rate of change $ \frac{1}{r} \frac{dr}{dt} $ (in s$^{-1}$) is given by:
- JEE Advanced - 2025
- Physics
- Gravitation
- If the weight of an object is 200 Newtons, then the weight of the object at the midpoint between the Earth's surface and the Earth's center will be:
- UPCATET - 2025
- Physics
- Gravitation
View More Questions
Questions Asked in JEE Advanced exam
- Let $ \mathbb{R} $ denote the set of all real numbers. Let $ a_i, b_i \in \mathbb{R} $ for $ i \in \{1, 2, 3\} $.
Define the functions $ f: \mathbb{R} \to \mathbb{R},\ g: \mathbb{R} \to \mathbb{R},\ h: \mathbb{R} \to \mathbb{R} $ by:
$$ f(x) = a_1 + 10x + a_2x^2 + a_3x^3 + x^4,\quad g(x) = b_1 + 3x + b_2x^2 + b_3x^3 + x^4, $$ $$ h(x) = f(x+1) - g(x+2) $$ If $ f(x) \ne g(x) $ for every $ x \in \mathbb{R} $, then the coefficient of $ x^3 $ in $ h(x) $ is:- JEE Advanced - 2025
- Polynomials
Let $ y(x) $ be the solution of the differential equation $$ x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e}, $$ satisfying $ y(1) = 0 $. Then the value of $ 2 \cdot \frac{(y(e))^2}{y(e^2)} $ is ________.
- JEE Advanced - 2025
- Differential equations
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
- JEE Advanced - 2025
- Coordinate Geometry
- A projectile is thrown at an angle of \(60^\circ\) with the horizontal. Initial speed is \(270\, \text{m/s}\). A linear drag force \(F = -CV\) acts on the body. Find the horizontal displacement till \(t = 2\) seconds. Given \(C = 0.1\, \text{s}^{-1}\).
- JEE Advanced - 2025
- Projectile motion
- A single slit diffraction experiment is performed to determine the slit width using the equation, $ \dfrac{b d}{D} = m \lambda $, where $ b $ is the slit width, $ D $ the shortest distance between the slit and the screen, $ d $ the distance between the $ m^{\text{th}} $ diffraction maximum and the central maximum, and $ \lambda $ is the wavelength. $ D $ and $ d $ are measured with scales of least count of 1 cm and 1 mm, respectively. The values of $ \lambda $ and $ m $ are known precisely to be $ 600 \, \text{nm} $ and 3, respectively. The absolute error (in $ \mu\text{m} $) in the value of $ b $ estimated using the diffraction maximum that occurs for $ m = 3 $ with $ d = 5 \, \text{mm} $ and $ D = 1 \, \text{m} $ is ___.
- JEE Advanced - 2025
- Error Propagation
View More Questions
Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].