Question 174: Work, Energy and Power & Circular Motion

Question

A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s
2
)

Accepted Answer

Let the speed of bob at lowest position be v
1
and at the highest position be v
2
.
Maximum tension is at lowest position and minimum tension is at the highest position. Now, using, conservation of mechanical energy,
$${1 \over 2}mv_1^2 = {1 \over 2}mv_2^2 + mg2l$$
$$ \Rightarrow {v_1}^2 = {v_2}^2 + 4gl$$ ..........(1)
Now, $${T_{\max }} - mg = {{mv_1^2} \over l}$$
$$ \Rightarrow {T_{\max }} = mg + {{mv_1^2} \over l}$$
& $${T_{\min }} + mg = {{mv_2^2} \over l}$$
$$ \Rightarrow {T_{\min }} = {{mv_2^2} \over l} - mg$$
$${{{T_{\max }}} \over {{T_{\min }}}} = {5 \over 1}$$
$$ \Rightarrow {{mg + {{mv_1^2} \over l}} \over {{{mv_2^2} \over l} - mg}} = {5 \over 1}$$
$$ \Rightarrow mg + {{mv_1^2} \over l} = \left[ {{{mv_2^2} \over l} - mg} ight]5$$
$$ \Rightarrow mg + {m \over l}\left[ {v_2^2 + 4gl} ight] = {{5mv_2^2} \over l} - 5mg$$
$$ \Rightarrow mg + {{mv_2^2} \over l} + 4mg = {{5mv_2^2} \over l} - 5mg$$
$$ \Rightarrow 10mg = {{4mv_2^2} \over l}$$
$${v_2}^2 = {{10 	imes 10 	imes 1} \over 4}$$
$$ \Rightarrow {v_2}^2 = 25 \Rightarrow {v_2} = 5$$ m/s
Thus, velocity of bob at highest position 5 m/s.

Work, Energy and Power & Circular Motion

A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ________ m/s. (Take g = 10 m/s 2 )

Integer / Numeric Answer Type