A ball with a radius of 015 m and mass of 3 kg starts from...

A ball with a radius of 0.15 m and mass of 3 kg starts from rest and rolls down a ramp with a height of 14 m and angle of 40 degrees. What is the linear velocity in m/s when the ball reaches the bottom of the ramp?

Transcript

00:01 In this question we have given that a ball with a radius of 0 .1 meter and mass 3 kilograms starts from the rest and rolls down a ramp with a height of 14 meter at an angle of 40 degrees then we have to find the linear velocity when the ball reaches to the bottom of the ramp so we have given data for this question that is mass of the ball m is 3 kilogram we have given its radius which is r that is 0 .15 meter we have given the height of the ramp h is 14 meter and the inclination theta is 40 degree so if we draw the diagram for this question so let this is the ramp here with an inclination of theta and there is a ball which slides down and it reaches to the bottom of this ramp this is the final position of the ball this is the initial position so we can see that let this position is first and this is the second position.

01:12 So as it rolls down, so it has angular velocity and linear velocity.

01:18 So let at this point its angular velocity is omega and it has a linear velocity that is v.

01:26 So we have to find this linear velocity.

01:29 So we can apply conservation of energy at point one end.

01:47 This is the height of the ramp that is h so we can say that at this point it is in initially in rest so it has only potential energy so we can say at kinetic energy at point 1 plus potential energy at point 1 that is equal to kinetic energy at 2 plus potential energy at point 2 so the kinetic energy at 0 .1 is 0 plus potential energy is m g into height h and when it reaches at the bottom there is no potential energy because this is the reference so it has only kinetic energy and the kinetic energy is due to the translation that is half mv square plus kinetic energy due to the rotation is half i omega square so this is zero now we know that i4 ball is 2 by 5 mr square because ball is a solid sphere so its moment of inertia is 2 by 5 mr square so this is m gh that is equal to half mv square plus half 2 by 5 m r square into omega is v by r square because for pure rolling we know that v is r into omega now simplify this equation and we can say this is m gh is equal to this is half mv square plus this is 1 by 5 mv square m will cancel out and finally we can say that this is g h h h is h h h that is equal to 7 by 10 v square...

Question

A ball with a radius of 0.15 m and mass of 3 kg starts from rest and rolls down a ramp with a height of 14 m and angle of 40 degrees. What is the linear velocity in m/s when the ball reaches the bottom of the ramp?

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