Question

Arm OA rotates counterclockwise with a constant angular velocity of ω = 5 rad/s. As the arm passes the horizontal position, a 6 kg ball is placed at the end of the arm. As the arm moves upward, the ball begins to roll, with negligible rolling resistance, towards the pivot O. It is noted that at θ = 30 ◦ , the ball is 0.9 meters from the pivot and moving towards O along the length of the arm. The ball moves with a speed of 0.4 m/s along the bar. What is the normal force that the arm applies to the ball at this instant? Please, indicate which principle you are applying and explain why.

Answer 1

26.92 N

Step-by-step explanation:

The normal reaction of the ball is due to two force component acting on it.

1. The normal reaction as a result of the weight of the ball
2. The normal reaction due to the component of the acceleration of the ball with the rod.

However ; the acceleration is in polar coordinate which is given by the relation:

`a^ { ^ \to} = (r- r \omega^2) \hat {e_r} + ( r \theta + 2 r \omega ) \hat {e_ \theta}`

`a_(\theta) = r \theta + 2 r \omega`

Given that :

ω = 5 rad/s

mass m = 6 kg

θ = 30 ◦

r = 0.9 m

speed v = 0.4 m/s

`a_(\theta) = 0 + 2(-0.4)*5`

`a_(\theta)= -4 \ m/s`

The normal force reaction (N) that the arm applies to the ball at this instant is :

N = mg cos θ +
`ma_(\theta)`

N = (6 × 9.8× cos 30) + (6 ×(-4))

N = 26.92 N