A 5.83-kg ball hangs from the top of a vertical pole by a 2.03-m-log string The

Question

A 5.83-kg ball hangs from the top of a vertical pole by a 2.03-m-log string The ball is struck, causing it to revolve around the pole at a speed of 4.89 m/s in a horizontal carcle with the string remaining taut Calculate the angle, between 0° and 90", that the string maker; with the pole. Take g # 9.81 me? Number 41.22 5.83 kg What is the tension of the string? Number 76.04 There is a hint available! View the hint by clicking on the bottom divider bar, Click on the divider bar again to hide the hint.

Explanation / Answer

along vertical

T*costheta = m*g...........(1)

along horizontal

T*sintheta = m*v^2/r.........(2)

r = radius of circle = L*sintheta

from 1 & 2

sintheta/costheta = v^2/(r*g)

sintheta/costheta = v^2/(L*sintheta*g)

(sintheta)^2/costheta = v^2/(L*g)

sintheta^2 = 1- (costheta)^2

(1- (costheta)^2)/costheta = v^2/(Lg)

costheta = x

(1-x^2)/x = v^2/(Lg)

(1-x^2)/x = 4.89^2/(2.03*9.8)

x = 0.566

costheta = 0.566

theta = cos^-1(0.566) = 55.3 degrees <<<------ANSWER

==================

from 1

T*costheta = m*g

T*cos55.53 = 5.83*9.8

T = 101 N <<<<-----ANSWER

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