Velocity of centre of mass (rolling w/o slipping)

[redtrebor]

Find the velocity of the centre of mass of a glass marble which rolls from rest down a
gentle ramp inclined at an angle of 10o
to the horizontal, without slipping, after it has
travelled for a distance of 2.3 m.



Moment of inertia of sphere: 2/5MR^2



Past paper question that I'm really struggling with.

 

[BvU]

Hi treb, and welcome to PF.
How did you bypass the template ? Folks at PF don't enforce it strictly to pester you! Use it and you'll get answers.

 

[redtrebor]

Hi, that was as close to the template as i could manage I'm afraid. I really don't know where to start with this question

 

[BvU]

I need a bit more to help you adequately. Ever heard of potential energy ? Can you connect rotation and velocity of center of mass with each other ?

 

[HalfLostAndSearching]

The velocity of the centre of mass can be calculated using the formula v = ωr, where v is the linear velocity, ω is the angular velocity, and r is the distance from the centre of mass to the point of rotation. In this case, the marble is rolling without slipping, so the angular velocity is equal to the linear velocity divided by the radius, ω = v/r.

To find the linear velocity, we can use the equation v^2 = u^2 + 2as, where u is the initial velocity (which is 0 in this case), a is the acceleration, and s is the distance travelled. In this scenario, the acceleration is due to gravity and can be calculated using a = gsinθ, where g is the acceleration due to gravity (9.8 m/s^2) and θ is the angle of the ramp (10 degrees).

Plugging in the values, we get v^2 = 0 + 2(9.8 m/s^2)(2.3 m)sin(10o) = 42.4 m^2/s^2. Taking the square root, we get v = 6.5 m/s.

Now, to find the angular velocity, we can use the formula ω = v/r. The radius of the marble can be calculated using the moment of inertia formula for a sphere, I = 2/5MR^2, where M is the mass of the marble and R is the radius. Rearranging for R, we get R = √(5I/2M). Plugging in the values of I (2/5MR^2) and M, we get R = √(5(2/5MR^2)/2M) = √(R^2) = R.

Therefore, the angular velocity is ω = v/R = 6.5 m/s / R.

To find the final velocity of the centre of mass, we can use the formula v = ωr, where r is the distance travelled (2.3 m). Plugging in the values, we get v = (6.5 m/s)(2.3 m) = 14.95 m/s.

In conclusion, the velocity of the centre of mass of the glass marble rolling without slipping down a ramp inclined at 10 degrees after travelling a distance of 2.3