Question

Q]. A body weighing 0.5 kg tied to a string is projected with a velocity of 10 m/s. Thebody starts whirling in a vertical circle. If the radius of the circle is 0.8 m, find the tension in the string when the body is at the top of the circle​

Answer 1

${ \underline{ \underline{ \huge{ \rm{ \green{QUESTION}}}}}}$

A body weighing 0.5 kg tied to a string is projected with a velocity of 10 m/s. The body starts whirling in a vertical circle . lf the radius of the circle is 0.8 m, find the tension in the string when the body is at the top of the circle ?

${ \underline{ \underline{ \huge{ \rm { \green{SOLUTION}}}}}}$

${ \dag{ \underline{ \underline{ \purple{ \tt{ \: \: \: given}}}}}} \\ { \dashrightarrow{ \blue{ \sf{ \: mass \: (m) = 0.5 \: kg}}}} \\ { \dashrightarrow{ \blue{ \sf{ \: initial \: velocity \: (Vo) = 10 \: m \: {sec}^{ - 1} }}}} \\ { \dashrightarrow{ \blue{ \sf{ \: length \: of \: string \: ( \: l \: ) = 0.8 \: m}}}}$

${ \dag{ \underline{ \underline{ \sf{ \purple{ \: \: \: to \: find}}}}}} \\ \\{ \sf{ tension \: in \: the \: string }} \\{ \sf { when \: the \: body \: is \: at \: the \: top \: of \: vertical \: circle}}$

${ \dag{ \underline{ \underline{ \purple{ \sf{ \: \: formula \: used}}}}}} \\ \\{ \sf{ tension \: in \: the \: string \:when \: the \: body}} \\{ \sf{is \: at \: the \: top \: of \: vertical \: circle}} \\ \\ { \boxed{ \boxed{ \sf{ \green{T \: = m( \frac{ {Vo}^{2} }{l} - 5g)}}}}}$

${ : { \implies{ \blue{ \sf{ \: T \: = 0.5 \: ( \frac{ {(10)}^{2} }{0.8} - 5(10))} \:N \: }}}}$

${ : { \implies{ \blue{ \sf{T = 0.5( \frac{1000}{8} - 50)} \: N \: }}}}$

${ : { \implies{ \blue{ \sf{ \: T = 0.5( \frac{1000 - 400}{8})} \: N \: }}}}$

${ : { \implies{ \blue{ \sf{ \: T = \frac{0.5 \times 600}{8} \: N \: \: }}}}}$

${ \boxed{ \boxed{ \red{ \sf{T = 37.5 \:N \: }}}}}$

${ \underline{ \underline{ \huge{ \rm{ \green{CONCEPT}}}}}}$

refer to the attached images

Answer 2

Answer:

☆Concept:

▪︎during vertical circular motion of the object kinetic energy gets changed to potential and vice versa.

▪︎ Existing force,Tension will provide the centripetal force.

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》observing the object by its frame of reference,

it will experience the psuedo-force,centrifugal towards outside.

☆Solution:

let,

Tension at top= $\ T_{t}$,Tension at bottom= $\ T_{b}$

Velocity at top=$\ v_{t}$,Velocity at bottom=$\ v_{t}$

¤Energy conservation:

$\green{\bf{ KE_{i} +PE_{i} = KE_{f} +PE_{f}}}$

$\rightarrow{\sf{ \dfrac{1}{2}m(v_{b})^{2} + mgh=\dfrac{1}{2}m(v_{t})^{2} +mgh'}}$

$\rightarrow{\sf{ \dfrac{1}{2}m(v_{b})^{2} + mg(0)=\dfrac{1}{2}m(v_{t})^{2} +mg(2l)}}$

$\rightarrow{\sf{ \dfrac{1}{2}m(10)^{2} +0 =\dfrac{1}{2}m(v_{t})^{2} +mg(2l)}}$

$\rightarrow{\sf{ \dfrac{1}{2}(100)=\dfrac{1}{2}(v_{t})^{2} +g(2l)}}$

$\rightarrow{\sf{ 50=\dfrac{1}{2}(v_{t})^{2} +16}}$

$\implies{\bf{(v_{t})^{2}=68}}$

¤Centrifugal force :

》Tension at top:

$\bf{ T_{t}}$ +mg = $\bf{m \dfrac{(v_{t})^{2}}{r}}$

$\bf{ T_{t}}$ = $\bf{m \dfrac{(v_{t})^{2}}{r}}$ - mg

$\rightarrow{\sf{T_{t} =(0.5×68/0.8) - (0.5×10)}}$

$\rightarrow{\bf{ T_{t} = 42.5 - 5}}$

$\rightarrow{\bf{ T_{t} = 37.5}}$

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so tension at top = 37.5N

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♧hope it helps!♧