Question

A well-greased, essentially frictionless, metal bowl has the shape of a hemisphere ok radius 0.150 m. You place a pat of butter of mass 5.00 x 10 kg at the rim of the bowl and let it slide to the bottom of the bowl. What is the speed of the pat of butter when it reaches the bottom of the bowl? At the bottom of the bowl, what is the force that the bowl exerts on the pat of butter?

How does this force compare to the weight of the pat?

Answer 1

To find the speed of the pat of butter at the bottom of the bowl, use the conservation of energy principle and the equation v = √(2gh). For the force, apply Newton's second law with centripetal acceleration to find F = mv²/r. The force exerted by the bowl is greater than the weight of the pat due to centripetal force.

Step-by-step explanation:

Speed of the Pat of Butter and Force Exerted at the Bottom of the Bowl

To calculate the speed of the pat of butter when it reaches the bottom of the bowl, we can use the principle of conservation of energy. Initially, the pat of butter has only potential energy and no kinetic energy. When it slides to the bottom, all potential energy has been converted into kinetic energy because there are no other energy losses due to friction or air resistance in this idealized scenario. The conservation of energy can be written as:

PEinitial + KEinitial = PEfinal + KEfinal

PE(initial) = mgh, where m is mass (5.00 x 10 kg seems incorrect and may be a typo; assuming it is 5.00 kg instead), g is acceleration due to gravity (9.81 m/s2), and h is the height (equal to the radius r = 0.150 m). KE(initial) is zero since it starts from rest.

At the bottom of the bowl, PE(final) is zero because h=0, and KE(final) = 0.5 * m * v2, where v is the final velocity.

Therefore, mgh = 0.5 * m * v2 which simplifies to v = √(2gh). Substituting values, we find the speed of the pat of butter at the bottom of the bowl.

To calculate the force the bowl exerts on the pat of butter, we use Newton's second law, F=ma, where a is the centripetal acceleration required to keep the butter moving in a circle at the bottom of the bowl, which is v2/r. The force exerted by the bowl is therefore F = m * v2/r.

The force that the bowl exerts on the pat of butter at the bottom is equal to the normal force and will be greater than the weight of the pat (mg) due to the additional centripetal force required for circular motion.

Answer 2

The problem can be solved by applying the conservation of energy principle and calculating the normal force at the bottom of the bowl using centripetal force requirements. The butter's potential energy at the rim converts into kinetic energy at the bowl's bottom, and the force of the bowl on the butter at the bottom can be compared to the butter's weight.

Step-by-step explanation:

The question asks about the speed of a pat of butter as it slides down a frictionless hemispherical bowl and the force exerted by the bowl at the bottom. To find the speed at the bottom of the bowl, we use the conservation of energy principle. The butter's potential energy at the top is completely converted into kinetic energy at the bottom. Thus, we can use the formula mgh = 1/2 mv², where m is the mass of the butter, g is acceleration due to gravity (9.8 m/s²), h is the height of the bowl's rim from the bottom (equal to the radius, r), and v is the velocity.

Plugging the values into the formula: (5.00 x 10 kg)(9.8 m/s²)(0.150 m) = 1/2 (5.00 x 10 kg)v². Solving for v, we would get the speed of the butter at the bottom of the bowl.

When the butter reaches the bottom, the force that the bowl exerts on the butter is the normal force, which must equal the centripetal force required to maintain the butter's circular motion at the bottom of the bowl. This force is given by F = mv²/r. We calculate this using the mass of the butter and the velocity we obtained previously. To compare this force to the weight of the pat of butter, one can simply compare the magnitude of the normal force with the butter's weight (mass times gravity).