Translational velocity help

[blink1987]

Translational velocity help!

Homework Statement

A ball is on an incline with µ = 30±. Its initial height is 1.5 m. When
it reached the bottom of the incline, what is the ball’s translational (linear)
velocity? [Use conservation of energy to solve.]

Homework Equations

r= s$$\vartheta$$

The Attempt at a Solution

r= (1.5)sin(30)

I am confused with this problem because a radius was not given and I am not sure if I started it right...

 

[whybother]

I think you aren't going about the problem the way your teacher intended - Go with the hint, to use conservation of energy.

At the top of the ramp, the ball is not moving, so all the energy is in the form of potential energy. At the bottom of the ramp, all of the energy is now in kinetic energy. Your relevant equations should be the form of potential and kinetic energy knowing that Initial Energy = Final Energy here. They specify linear velocity too, to let you know that basically the rotation of the ball is to be ignored (ie. you don't really need to concern yourself with the radius).

 

[nlightner]

Hello! It seems like you are working on a physics problem involving an incline and conservation of energy. To find the translational velocity of the ball at the bottom of the incline, you will need to use the conservation of energy equation, which states that the total initial energy of the ball (potential energy + kinetic energy) is equal to the total final energy of the ball (potential energy + kinetic energy).

First, let's determine the total initial energy of the ball. The initial potential energy of the ball is given by mgh, where m is the mass of the ball, g is the acceleration due to gravity, and h is the initial height of the ball. In this case, m and g are not given, but we can assume that m is constant and g is equal to 9.8 m/s^2. So, the initial potential energy of the ball is mgh = m(9.8)(1.5) = 14.7m.

Next, we need to determine the total final energy of the ball. At the bottom of the incline, all of the initial potential energy of the ball is converted into kinetic energy, since the ball is no longer at a height. The final kinetic energy of the ball is given by (1/2)mv^2, where m is the mass of the ball and v is the final translational velocity.

Now, we can set up the conservation of energy equation:

Total initial energy = Total final energy
mgh = (1/2)mv^2

Substituting in the values we found earlier for the initial potential energy and solving for v, we get:

14.7m = (1/2)mv^2
29.4 = v^2
v = √29.4 ≈ 5.42 m/s

So, the translational velocity of the ball at the bottom of the incline is approximately 5.42 m/s. I hope this helps! Let me know if you have any further questions.