Question

a 1kg ball is bening pushed by the rod to move in horizontal groved smooth slot if it startes from angle teta = zero degree . determaine the force the rod exertes on the ball at teta is =15 dgree if ai this instant the rod moves at angular speed of teta = 1 rad per sec end with angular acceleration theta = 2 rad persec and square the ball is only in contact with the outer side of the slot​

Answer 1

The force the rod exerts on the ball at the given angle is determined as 3.94 N.

Force exerted on the rod by the ball

The force exerted is calculated as follows;

F = ma

F = mv²/r

F = mω²r

where;

• m is mass of the ball
• ω is angular speed of the ball
• r is radius of the path

r = 2cosθ

Angular speed when the ball moves 15 degrees

ωf² = ωi² + 2αθ

where;

• θ is angular displacement in radians, 15⁰ = 15 x π/180 rad

ωf² = (1)² + 2(2)(15 x π/180)

ωf² = 2.04

ωf = √2.04

ωf = 1.428 rad/s

F = mω²(2cosθ)

F = (1)(1.428)²(2 x cos15)

F = 3.94 N

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