Rolling Ball on a Horizontal Bar with Angular Velocity ω

[m_34n814]

bar (length L) is turned with angular velocity ω, on a horizontal level. Αt length of the bar, there is a ball (mass m) which rolls (We suppose that there is no friction force). The ball begins from constant utmost the bar with initial velocity u0. When the ball reaches in the L?

Can you help me!?
Thank you!

 

[DaveC426913]

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[m_34n814]

First of all, according to the no Inertial Reference Frame there will be 2 Forces (the once is Coriolis and the other once is Centrifugal).
We can write:
xdx^2/dt^2=-2w*v=-2w*v(z*y)=2w*v*x
or d^2x/dt^2=2*w*v
where v is the velocity of the ball.
I think that I have to write v connection x, and then I can solve this differential equation…
Then I will solve the equation x=L to find t.

My problem is how can I connect v and x.

I think that my problem is the direction of these 2 Forces
In fact, I think that I haven’t understand how the ball will move.

 

[tiny-tim]

hi m_34n814!

(have an omega: ω and try using the X² icon just above the Reply box )

the https://www.physicsforums.com/showthread.php?t=456090&nojs=1#goto_threadtools" is always perpendicular to v_rel, the velocity relative to the moving frame …

so it's tangential, and has no effect on the radial motion …

you can ignore it ​

 

[PJC01]

I would be happy to help you understand the situation of the rolling ball on a horizontal bar with angular velocity ω. First, we need to understand the concept of angular velocity, which is the rate at which an object rotates around a fixed point. In this case, the bar is the fixed point and it is being turned with a constant angular velocity ω.

Next, we have a ball located at the end of the bar, and it has a mass of m. Since there is no friction force, we can assume that the ball will roll without any resistance. The initial velocity of the ball, u0, is also given.

Now, as the bar continues to turn with angular velocity ω, the ball will also rotate with the same angular velocity. This means that the ball will have a tangential velocity at any point along the bar, which is equal to the product of its angular velocity ω and the distance from the fixed point (L in this case). This tangential velocity will cause the ball to move along the bar.

As for the question of when the ball reaches L, we can use the equation of motion for rotational motion to determine the time it takes for the ball to reach L. This equation is:

θ = ωt + 1/2αt^2

Where θ is the angular displacement, ω is the angular velocity, α is the angular acceleration (which is zero in this case), and t is the time.

Since we know that the initial angular displacement is zero (the ball starts at the end of the bar), and the final angular displacement is L (the end of the bar), we can rearrange the equation to solve for t:

t = √(2L/ω)

Therefore, it will take the ball √(2L/ω) seconds to reach L. I hope this explanation helps you understand the situation better. If you have any further questions, please feel free to ask.