Conservation of Energy question

In summary, the conversation discusses a scenario where a particle of mass m is released from rest into a hemispherical bowl of mass M on a table with a frictionless inside surface but a coefficient of friction between the bottom of the bowl and the table. The goal is to find the largest value of m/M for which the bowl never slides on the table. The conversation suggests using energy conservation and Newton's equations to solve for the unknowns and using the inequality Fs < (static friction coef.) * N_M to find the desired value of m. Fig. 5.38 is referenced for further understanding of the scenario.
  • #1
zheng89120
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0

Homework Statement



A hemispherical Bowl of mass M rests on a table. The inside surface of the bown is frictionless, while the coefficient of friction between the bottom of the bowl and the table is u = 1. A particle of mass m is released from rest at the top of the bowl and slides down into it, as shown in Fig. 5.38. What is the largest value of m/M for which the bowl never slides on the table? Hint: the angle you're concerned with is not 45.

Homework Equations



Energy conservation, maybe Newton's equations

The Attempt at a Solution



some form of sigma E = sigma E, not sure
 
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  • #2
Please send Fig 5.38
And what angle is that in the hint, maybe the Fig will help.

Considering that you have Friction and falling bodies start by writing the relevant equations of these and look for how many unknowns you have.
 
  • #4
First, only consider the situation where the bowl stays at rest. In this case, m performs circular motion. Find the normal force N_m on m in terms of its speed v and its angle position theta. The speed v can be found using the law of energy conservation.

Next, draw a force diagram for the bowl M. Find the normal force N_M acting on it and the horizontal force F that m exerts on M. F should be equal to the static friction force Fs in order to keep M remain resting.

Finally apply the equation: Fs < (static friction coef.) * N_M
Fs and N_M are expressed in terms of m, M and theta. Plug them into the above inequality and find m which satisfies the inequality for all values of theta in the range [0;pi] (or maybe [-pi/2 ; pi/2] depending which angle you choose).
 
  • #5
how to incorporate the friction force into this equation.

I would approach this problem by first considering the conservation of energy principle. This states that the total energy of a closed system, such as the bowl and particle in this scenario, remains constant over time. In this case, the energy of the system is equal to the sum of the potential energy and kinetic energy.

At the top of the bowl, the particle has only potential energy, which is equal to its mass (m) multiplied by the acceleration due to gravity (g) and the height of the bowl (h). As it slides down the bowl, this potential energy is converted into kinetic energy, which is equal to half of the particle's mass multiplied by its velocity squared.

At the point where the particle reaches the bottom of the bowl, it will have a velocity that is equal to the square root of 2gh. This velocity will then be used to calculate the kinetic energy of the system. However, at this point, the bowl will also start to slide on the table due to the friction force. This means that some of the kinetic energy of the system will be converted into work done by the friction force, which is equal to the coefficient of friction (u) multiplied by the normal force (mg) and the distance the bowl slides (h).

To find the largest value of m/M for which the bowl never slides on the table, we can set the kinetic energy of the system at the bottom of the bowl equal to the work done by the friction force:

1/2 * m * (sqrt(2gh))^2 = u * mg * h

Solving for m/M, we get:

m/M = (2u * h)/g

Since we know that the angle we are concerned with is not 45 degrees, we can use trigonometry to find the height (h) of the bowl in terms of its radius (R) and the angle (θ) between the vertical and the tangent line at the bottom of the bowl:

h = R * (1 - cos(θ))

Substituting this into our equation for m/M, we get:

m/M = (2u * R * (1 - cos(θ)))/g

Therefore, the largest value of m/M for which the bowl never slides on the table is when the angle (θ) is equal to 90 degrees, or when the bowl is completely vertical. In this case, the maximum
 

1. 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 transformed from one form to another.

2. Why is the conservation of energy important?

The conservation of energy is important because it is a fundamental law of physics that helps us understand and predict the behavior of energy in various systems. It also allows us to use energy more efficiently and sustainably.

3. How does the conservation of energy apply to everyday life?

The conservation of energy applies to everyday life in many ways. For example, when we turn on a light switch, the electrical energy is converted to light and heat energy. When we ride a bike, our muscles convert chemical energy into mechanical energy. Understanding the conservation of energy can also help us make informed choices about energy usage and conservation.

4. Are there any exceptions to the law of conservation of energy?

While the law of conservation of energy holds true in most cases, there are a few exceptions. In nuclear reactions, a small amount of mass can be converted into a large amount of energy, according to Einstein's famous equation E=mc². Additionally, in quantum mechanics, energy can fluctuate or appear out of nowhere for short periods of time, but it still adheres to the overall principle of conservation.

5. How does the conservation of energy relate to other laws of physics?

The conservation of energy is closely related to the first law of thermodynamics, which states that energy cannot be created or destroyed in a closed system. It also plays a role in many other laws and principles, such as the law of conservation of momentum and the principle of least action.

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