Answer Work & Kinetic Energy for Mass m Whirling on Table

In summary, a mass on a frictionless table is held to a circular motion by a string passing through a hole in the table. When the radius of the circle changes from ##l_1## to ##l_2## due to the string being slowly pulled, the work done in pulling the string is equal to the increase in kinetic energy of the mass. This is because the rope tension, which is a central force, does work on the mass from ##l_1## to ##l_2##, resulting in a change in kinetic energy. This can be expressed as ##W_{21} = \frac{1}{2} m (v^2(l_2) - v^2(l_1)) = \
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Homework Statement


Mass m whirls on a frictionless table, held to a circular motion by a string which passes through a hole in the table. The string is slowly pulled through the hole so that the radius of the circle changes from ##l_1## to ## l_2## Show that the work done in pulling the string equals the increase in kinetic energy of the mass

Homework Equations


Work and energy

The Attempt at a Solution


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I guess that the statement "the string is slowly pulled" means that the distance of the mass from the hole cannot change much in a short time, so I assume that ##\dot r = v_r = cst## and ##\ddot r = 0 ##.

By Newton second law in the radial direction:
## \vec T = m\ \vec a_r = m ( \ddot r - r {\dot \theta} ^2)\ \hat r= - mr {\dot \theta} ^2\hat r ##

There no tangential acceleration because there is no force in that direction so :
## 0 = a_\theta = r\ddot \theta + 2 \dot r \dot \theta = \frac{1}{r}\frac{d}{dt}(r^2\dot\theta) ##

So the quantity ## r^2\dot\theta ## is constant, and ##{\dot \theta} ^2 = \frac{c^2}{r^4}## where ##c ## is constant, and rope tension is now ##\vec T = -mc^2\frac{\hat r}{r^3}##

Since rope tension is a central force, its work from ##l_1## to ##l_2## is :
##W_{21} = \int_{l_1}^{l_2} T(r) dr = -mc^2 \int_{l_1}^{l_2} \frac{1}{r^3} dr = \frac{1}{2} m c^2 (\frac{1}{l_2^2} - \frac{1}{l_1^2})##

Because ## c = r^2\dot\theta = r v_\theta ##, and because ## \dot r = v_r ## is constant, then ##
\begin{align}
W_{21} =& \frac{1}{2} m ( v_\theta^2(l_2) - v_\theta^2(l_1)) \\
= & \frac{1}{2} m ( v_r^2(l_2) + v_\theta^2(l_2) - (v_r^2(l_1) +v_\theta^2(l_1))) \\
= & \frac{1}{2} m ( v^2(l_2) - v^2(l_1)) \\
=& \triangle K
\end{align}
##

Is that correct ?
 
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YES! :)
 

1. What is the difference between work and kinetic energy?

Work is the measure of energy transferred to or from an object by a force. Kinetic energy is the energy an object possesses due to its motion. In other words, work is the act of applying a force to move an object, while kinetic energy is the result of that object's movement.

2. How is work calculated for an object whirling on a table?

The work done on an object whirling on a table can be calculated by multiplying the force applied to the object by the distance it moves in the direction of the force. This is represented by the equation W = F x d, where W is work, F is force, and d is distance.

3. What is the formula for calculating kinetic energy?

Kinetic energy can be calculated using the equation KE = 1/2 x m x v^2, where KE is kinetic energy, m is the mass of the object, and v is the velocity of the object.

4. How does mass affect work and kinetic energy for an object whirling on a table?

The mass of an object affects both work and kinetic energy for an object whirling on a table. Greater mass requires more work to be done to achieve the same change in velocity. The greater the mass, the more kinetic energy the object possesses at a given velocity.

5. Can work and kinetic energy be converted into one another?

Yes, work and kinetic energy can be converted into one another. When work is done on an object, it gains kinetic energy. On the other hand, when an object is stopped by a force, its kinetic energy is converted into work done by the force.

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