Question 321: Work, Energy and Power & Circular Motion

Question

A ball moves over a fixed track as shown in the figure. From $A$ to $B$, the ball rolls without slipping. Surface $B C$ is frictionless. $K_A, K_B$ and $K_c$ are kinetic energies of the ball at $A, B$ and C , respectively. Then

Accepted Answer

The work done by the friction force is zero in rolling without slipping. The surface BC is frictionless. Thus mechanical energy of the balls is conserved.
i.e., it is same at point $\mathrm{A}, \mathrm{B}$ and C .
If $K_A, K_B$ and $K_C$ represent the total oenergies (transtational plus rotational)
Then, $\mathrm{K}_{\mathrm{A}}+\mathrm{mgh}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}+\mathrm{mgh}_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$
The equation (i) gives $K_B>K_C$ and $K_B>K_A$ irrespective of the relation between $\mathrm{h}_{\mathrm{A}}$ and $\mathrm{h}_{\mathrm{C}}$.
Let us analyses the problem a little more.
Let $K_B{ }^{	ext {tans }}$ and $K_C{ }^{	ext {trans }}$ are translational kinetics energies and $K_B{ }^{	ext {rot }}$ and $K_C{ }^{	ext {rot }}$ rotational kinetic energies of balls.
The equation (i) can be written as
$$ \begin{aligned} \mathrm{K}_{\mathrm{A}}+m g h_A & =\left(\mathrm{K}_{\mathrm{B}}ight)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{B}}ight)_{\mathrm{rot}} \ & =\left(\mathrm{K}_{\mathrm{C}}ight)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{C}}ight)_{\mathrm{rot}}+m g h_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $$
Track $B C$ is frictionless and there is no torque on the ball about its centre of mass. Hence, its angular acceleration is zero, i.e., it moves with a constant angular velocity, which implies.
$$ \left(\mathrm{K}_{\mathrm{B}}ight)_{\mathrm{rot}}=\left(\mathrm{K}_{\mathrm{C}}ight)_{\mathrm{rot}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$$
The equation (iii) and (iv) gives $\left(\mathrm{K}_{\mathrm{B}}ight)_{	ext {trans }}>\left(\mathrm{K}_{\mathrm{C}}ight)_{	ext {trans }}$.

Answer

$h_{\mathrm{A}}>h_{\mathrm{c}^{\prime}} \mathrm{K}_{\mathrm{B}}>\mathrm{K}_{\mathrm{C}}$.

Answer

$h_{\mathrm{A}}>h_{\mathrm{c}} ; \mathrm{K}_{\mathrm{C}}>\mathrm{K}_{\mathrm{A}}$.

Answer

$h_{\mathrm{A}}=h_{\mathrm{c}} ; \mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}$.

Answer

$h_{\mathrm{A}} < h_{\mathrm{c}} ; \mathrm{K}_{\mathrm{B}}>\mathrm{K}_{\mathrm{C}}$.

Work, Energy and Power & Circular Motion

A ball moves over a fixed track as shown in the figure. From A to B, the ball rolls without slipping. Surface B C is frictionless. K_A, K_B and K_c are kinetic energies of the ball at A, B and C , respectively. Then

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