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Black_prince [1.1K]

2 years ago

12

A small ball is attached to the lower end of a 0.800-m-long string, and the other end of the string is tied to a horizontal rod.

The string makes a constant angle of 61.0 ∘ with the vertical as the ball moves at a constant speed in a horizontal circle.If it takes the ball 1.25 s to complete one revolution, what is the magnitude of the radial acceleration of the ball?

1 answer:

mestny [16]2 years ago

7 0

Answer:

a = 17.68 m/s²

Explanation:

given,

length of the string, L = 0.8 m

angle made with vertical, θ = 61°

time to complete 1 rev, t = 1.25 s

radial acceleration = ?

first we have to calculate the radius of the circle

R = L sin θ

R = 0.8 x sin 61°

R = 0.7 m

now, calculating at the angular velocity

$\omega =\dfrac{2\pi}{T}$

$\omega =\dfrac{2\pi}{1.25}$

ω = 5.026 rad/s

now, radial acceleration

a = r ω²

a = 0.7 x 5.026²

a = 17.68 m/s²

hence, the radial acceleration of the ball is equal to 17.68 rad/s²

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