21. Please show the solution in detail. A solid ball of mass 1.6 kg and radius 0.5 cm starts from rest then rolls without slipping down a long ramp themakes an angle 23 degrees with the horizontal. At what height above the bottom of the ramp must the ball bereleased if it is to have a speed of 3 m/s at the bottom?

Mass of the solid is given as, m = 1.6 kg Radius of the ball is given as, r = 0.5 cm Initially the ball is at rest, then the initial velocity is given as, vi = 0 Final velocity of the solid is given as, vf = 3 m/s Angle of inclination is given as, θ = 23ο TO DETERMINE: (a) The height (h) above the bottom of the ramp where the ball must be released to have a speed of 3 m/s at the bottom. From the given data, let us construct a diagram: For rotational motion, we know, Initial potential energy + Final Potential energy = Initial Kinetic energy + Final Kinetic energy Ui + Uf = Ki + Kf -------- (1) Where, Initial Potential energy, Ui = mgh ------- (2) Where, m is the mass of the object, m = 1.6 kg g is the gravitation energy, g = 9.8 m/s h is the height of the ramp. We don't have final potential energy since the objects starts to rotate on the ramp, Uf = 0 ----- (3) We don't have initial kinetic energy since the object is at rest, vi = 0, therefore, Ki = 0 --------- (4) Final Kinetic energy, Kf = Ktrans + Krot Ktrans is the translational kinetic energy, Ktrans = 12mv2 Krot is the rotational kinetic energy, Krot = 12Iω2 Where, I is the moment of inertia of the solid ball, I = 25mr2 ω is the angular velocity, ω = vr Here, both translational and rotational kinetic energy is present because the ball rolls down and continues to move on the horizontal floor. Therefore, we have Translational Kinetic energy and the ball continues to move in rotational motion only, therefore we have the Rotational Kinetic energy. Therefore, the final kinetic energy is given as, Kf = 12mv2 + 12Iω2 Kf = 12mv2 + 12×25mr2vr2 Kf =12mv2 + 15mv2 Kf = 710mv2 --------- (5)To calculate the height, We have, Ui + Uf = Ki + Kf Substituting the values of equation (2), (3), (4), and (5) in above equation, mgh + 0 = 0 + 710mv2 mgh = 710mv2 Rearranging the equation to calculate the height of ramp, h = 710v2g h = 710×(3)29.8 m h = 7×910×9.8 m h = 6398 m Therefore, the height of ramp is calculated as, h = 0.643 m (or) h = 64.3 cm(a) The height (h) above the bottom of the ramp where the ball must be released to have a speed of 3 m/s at the bottom is calculated as, h = 0.643 m (or) h = 64.3 cm

Answered: A solid ball of mass 1.6 kg and radius…

A solid ball of mass 1.6 kg and radius 0.5 cm starts from rest then rolls without slipping down a long ramp the makes an angle 23 degrees with the horizontal. At what height above the bottom of the ramp must the ball be released if it is to have a speed of 3 m/s at the bottom?