Derive an expression for linear velocity at lowest point and at highest point for particle revolving in vertical circular motion

Answers 1

[tex]\huge{\underline{\underline{\mathfrak{Answer :}}}}[/tex]

Let,

  • [tex]\sf{T_{L}}[/tex] = Tension at the lowest point.
  • [tex]\sf{T_{H}}[/tex] = Tension at the highest point.
  • [tex]\sf{V_{L}}[/tex] = Velocity at the lowest point.
  • [tex]\sf{V_{H}}[/tex] = Velocity at the highest point.

________________________

At lowest point L,

[tex]\sf{\rightarrow T_L = \dfrac{m {v_L}^2}{r} + mg}[/tex] _______(1)

At highest point H,

[tex]\sf{\rightarrow T_L = \dfrac{m {v_H}^2}{r} - mg}[/tex] _______(2)

________________________

Subtracting (1) by (2).

[tex]\sf{\rightarrow T_L - T_H = \dfrac{{mv_L}^2}{r} + mg - (\dfrac{{mv_H}^2}{r} - mg)}[/tex]

[tex]\sf{\rightarrow T_L - T_H = \dfrac{m}{r} ({v_L}^2 - {v_H}^2 + 2mg})[/tex] ______(3)

________________________

By law of conservation of energy,

(P.E + K.E) at L = (P.E + K.E) at H

[tex]\sf{ \therefore 0 + \dfrac{1}{2} {mv_L}^2 = mg.2r + \dfrac{1}{2} {mv_H}^2}[/tex]

[tex]\sf{ \therefore \dfrac{1}{2}m ({v_L}^2 - {v_H}^2) = mg.2r}[/tex]

[tex]\sf{ \therefore {v_L}^2 - {v_H}^2 = 4gr}[/tex]_______(4)

________________________

From equation (3) and (4),

[tex]\sf{\rightarrow T_L - T_H = \dfrac{m}{r}(4gr) + 2mg}[/tex]

[tex]\sf{\rightarrow T_L - T_H = 4mg + 2mg}[/tex]

[tex]\bf{ \therefore T_L - T_H = 6mg}[/tex]

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Akshat is a class 10 student. He went to his grandparent's home in a village. His grandfather took him to the bank of a nearby river. Akshat was very happy to see the pollution-free village environment. He was standing on the bank of the river observed that the angle of elevation of the top of a tree standing on the opposite bank was 60°. When he moved 30 metres away from the bank, he found the angle of elevation to be 30° .Find the height of the tree.Find the width of the river. [Take √3= 1.732.]​

Answer:

THE ANSWER IS

HEIGHT OF THE TREE= 26m

WIDTH OF THE RIVER= 15m

Step-by-step explanation:

let the width of the river be x and the height of the tree be h.

h/30+x= 1/root3--------(1)

h/x=root3-------(2)

after solving both equations we get,

30+x=3x

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x=15  (width of the river)

after substituting the value of x in equation 2 we get h= 25.96m which is equal to 26m.

Answer:

height of the tree is 15m

width of the river is 25.98

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