Ball of mass rolling down an incline

In summary, a ball of mass 2.80 kg and radius 0.148 m is released from rest on a plane inclined at an angle θ = 38.0° with respect to the horizontal. After rolling a distance d=1.55 m without slipping, the ball's speed is given by the equation PE0 = KEf + KErf. The initial kinetic energy is 0 and the initial rotational kinetic energy is also 0. The final potential energy is 0 since it was adjusted to be at ground level. The equation for finding the height of the initial potential energy is sin(theta) times d. It is recommended to double check calculations for any errors.
  • #1
delecticious
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Homework Statement


A ball of mass 2.80 kg and radius 0.148 m is released from rest on a plane inclined at an angle θ = 38.0° with respect to the horizontal. How fast is the ball moving (in m/s) after it has rolled a distance d=1.55 m? Assume that the ball rolls without slipping, and that its moment of inertia about its center of mass is 1.50E-2 kg·m2.
prob30_BallIncl.gif



Homework Equations



KE - kinetic energy
KEr - rotational kinetic energy
PE - potential kinetic energy

KE0 + KEr0 + PE0 = KEf + KErf + PEf

The Attempt at a Solution



I thought I had it right but it seems I went wrong somewhere and I'm assuming that it's probably where the potential energy is concerned. Since, the entire hypotenuse isn't given and all that is given is the distance d, I readjusted the axis for potential energy and made the point where the ball goes down to in the picture the ground level so that PE at that point equals 0. The initial KE is 0 as well as the initial rotational KE, and since I made the final PE equal 0 I got an equation PE0 = KEf + KErf. To find the height for the initial PE I did sin(theta) times the d. I guessing that that's the wrong thing to do. Anyone know the right way to go about it?
 
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  • #2
help?
 
  • #3
You have done everything correctly. Repeat the calculations to see if you've made any arithmetical mistake.
 

1. What factors affect the speed of a ball rolling down an incline?

The speed of a ball rolling down an incline is affected by several factors, including the angle of the incline, the mass of the ball, and the force of gravity. Other factors such as the surface of the incline and any external forces acting on the ball may also impact its speed.

2. How does the angle of the incline affect the motion of the rolling ball?

The angle of the incline plays a significant role in determining the motion of a rolling ball. A steeper incline will result in a faster acceleration of the ball due to the increased force of gravity pulling it downwards. A shallower incline will result in a slower acceleration, as the force of gravity is acting on the ball at a lesser angle.

3. Does the mass of the ball affect its acceleration down the incline?

Yes, the mass of the ball does affect its acceleration down the incline. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Therefore, a heavier ball will have a slower acceleration than a lighter ball down the same incline.

4. What is the difference between a ball rolling down an incline and a ball sliding down an incline?

When a ball rolls down an incline, it experiences both translational and rotational motion. In contrast, when a ball slides down an incline, it only experiences translational motion. This is because rolling motion involves the contact between the ball and the incline, causing it to rotate, while sliding motion does not.

5. How can the energy of a ball rolling down an incline be calculated?

The total energy of a rolling ball can be calculated by adding its kinetic energy (0.5mv^2) and its potential energy (mgh), where m is the mass of the ball, v is its velocity, and h is its height on the incline. This is known as the law of conservation of energy, which states that the total energy of a system remains constant. As the ball rolls down the incline, its potential energy decreases while its kinetic energy increases, but the total energy of the system remains the same.

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