A solid ball with a mass of 2.11 kg and a radius of 0.051 m...

A solid ball with a mass of 2.11 kg and a radius of 0.051 m starts from rest at the top of a 89.1° slope with a vertical height of 3.20 m. It proceeds to roll without slipping down the slope. What is the translational speed of the ball when it reaches the bottom? The acceleration of gravity is 9.81 m/s². Answer in units of m/s.

Transcript

00:01 In the problem, we are given a solid ball of mass 2 .11 kilograms and a radius of 0 .051 meters on top of an inclined plane with an angle of 89 .1 degrees and a height of 3 .20 meters.

00:24 Now, if the ball is to roll down the incline with zero initial velocity, we are asked to find what is the final velocity or the translational velocity.

00:38 Since there is no initial velocity, you could put here zero, and the gravitational acceleration is equal to 9 .81 meters per second square.

00:50 Now, we can solve this problem using the conservation of energy, where it states that the initial energy of the system is equal to the final energy of the system.

01:05 In this case, the energy can be divided into three terms, that is the potential energy plus the kinetic translational energy and the kinetic angular energy, or rotational kinetic energy.

01:27 So the left side of the equation is for the initial energies of the system, while the right side of the equation is for the final energies of the system.

01:41 Now we know that the gravitational potential energy is equal to mass, times gravity times the initial height, while the translational kinetic energy is equal to one -half mv squared.

01:59 And the rotational kinetic energy is one -half i omega -squared, where i is your moment of inertia and omega is the angular velocity.

02:12 Now the same would be here for the right side of the equation, but instead we use the final variables plus one half i omega l now we can substitute all the zero energies that we know for instance we know that the initial velocity is zero hence the kinetic energy translational and rotational and rotational would be zero so we're left with an initial energy from potential energy only on the right side of the equation we know that once the ball reaches the bottom of the inclined, the height would be zero.

03:01 So we could also remove this term here.

03:04 We are left with the final translational kinetic energy and the final rotational kinetic energy.

03:17 Now, since we are aware that the object rolling down the hill is a solid sphere, we could get its moment of inertia easy.

03:26 And the moment of inertia for a solid sphere or a solid ball is equal to 2 5th mass times radius squared.

03:39 Also, the relationship between linear velocity and angular velocity is simply omega is equal to v over r...

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