# Conservation of energy, centripetal acceleration, kinematics

• wcjy
In summary, the conversation discusses the correct formula for centripetal acceleration and clarifies the difference between the radius of curvature of the track and the radius of curvature of the ball's path. The correct formula is a = \frac{\left( \mathrm{speed \ of \ object} \right)^2}{\mathrm{radius \ of \ curvature \ of \ the \ object's \ path}} and the radius of curvature of the ball's path on the track is not the same as the radius of curvature of the track itself.

#### wcjy

Homework Statement
A small sphere of radius r_0 = 1.5 cm rolls without slipping on the track which is fixed on the
ground as shown in Figure 3(a). The radius of the track is R_0 = 26 cm. The sphere is released
from rest at height R_0 above the bottom of the track and rolls without slipping down the track.

(a) What is the angular velocity of the sphere when it leaves the track after passing through
an angle 125° as shown in Figure 3(a)

(b) What is the net acceleration of the sphere the moment before it leaves the track?

(c) At what distance D from the base of the track will the sphere hit the ground?
Relevant Equations
Conservation of Energy
a_c = v^2 / r

(a) Using COE,
$$mgh = 0.5mv^2 + 0.5I\omega^2$$
I solved it, where $$\omega = 112 rad/s$$

(b) This is the part where I have question or problem.
I saw my course mate working and he start of with finding centripetal acceleration.
$$a_c = \frac{v^2}{r} = \frac{(r_0\omega)^2}{R_0}$$

Why isn't it:
$$a_c = R_0\omega^2$$

Don't confuse the angular velocity of the sphere about its own centre with the angular velocity of the centre about the centre of curvature of the track.

wcjy
wcjy said:
(b) This is the part where I have question or problem.
I saw my course mate working and he start of with finding centripetal acceleration.
$$a_c = \frac{v^2}{r} = \frac{(r_0\omega)^2}{R_0}$$

Why isn't it:
$$a_c = R_0\omega^2$$
Neither is quite right. Although the approach by your course mate is closer.

Centripetal acceleration is
$$a = \frac{\left( \mathrm{speed \ of \ object} \right)^2}{\mathrm{radius \ of \ curvature \ of \ the \ object's \ path}}$$

The speed, with respect to the ground, of the ball is $r_0 \omega$ (where $r_0$ is the radius of the ball, and $\omega$ is the angular velocity of the ball around the ball's own center).

$R_0$ is the radius of curvature of the track, but that's not quite the same as the radius of curvature of the path of the ball as it rolls on the track. It's close, but not quite. Look at the diagram carefully. The radius of curvature of the ball's path on the track is what you are looking for. (Hint: the ball's radius is not zero.)

wcjy
Thanks for all your explanation. I think i get it. will be attempting the rest of the question and see if i face any difficulties. Thank You So Much!

## What is the law of conservation of energy?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. This means that the total amount of energy in a closed system remains constant over time.

## What is centripetal acceleration?

Centripetal acceleration is the acceleration that an object experiences when it moves in a circular path. It is always directed towards the center of the circle and is necessary to keep the object moving in a circular path.

## How is kinematics related to conservation of energy?

Kinematics is the study of motion without considering the forces that cause it. The conservation of energy is related to kinematics because it explains the relationship between an object's position, velocity, and acceleration, and how these factors affect the object's energy.

## What is the difference between potential and kinetic energy?

Potential energy is the energy that an object has due to its position or state, while kinetic energy is the energy that an object has due to its motion. Potential energy can be converted into kinetic energy and vice versa, but the total amount of energy remains constant.

## How can conservation of energy be applied in real-life scenarios?

Conservation of energy can be applied in various real-life scenarios, such as energy production and consumption, transportation, and even sports. For example, in a roller coaster, the potential energy at the top of the track is converted into kinetic energy as the coaster moves down the track, and the total energy remains constant. This principle is also used in renewable energy sources, such as solar and wind power, to convert energy from one form to another.

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