As no energy is lost/dissipated, total mechanical energy is conserved. Hence, the total energy is the same even when the ball reaches point B. However, as the height of the ball's position when it reaches point B is zero, all of its mechanical energy is stored as kinetic energy.
∑E = KE = KE(tran) + KE (rot) = 7/10 * Mv^2 = 437.6064J
7/10 * 4.8 * v^2 = 437.6064J
v^2 = 130.24
v = sqrt(130.24) = 11.4 m/s.
Therefore, the velocity of the ball when it reaches point B is 11.4 m/s.
Hi,
Thank you for your question. This is how I would solve these questions.
28 a)
The total energy of this ball as it rolls along the top of the hill is the gravitational potential energy and the kinetic energy of the ball.
Ug = mgh = 4.8 * 9.8 * 5.6 = 263.424J.
Remember, as the ball is rolling, there are both translational and rotational kinetic energy.
***The formula for moment of inertia of a solid sphere is I = 2/5 M * R^2
*** Angular velocity (ω) equals translational velocity (v) divided by radius (r).
∑KE = KE(tran) + KE(rot) = 1/2 Mv^2 + 1/2 I ω^2 = 1/2 Mv^2 + 1/2*2/5 M * R^2* v^2 / R^2
∑KE = 1/2 Mv^2 + 1/5 Mv^2 = 7/10 * Mv^2 = 7/10 * 4.8 * (7.2)^2 = 174.1824J
Therefore, the total energy is:
∑E = Ug + KE = 263.424J + 174.1824J = 437.6064J
28 c)
As no energy is lost/dissipated, total mechanical energy is conserved. Hence, the total energy is the same even when the ball reaches point B. However, as the height of the ball's position when it reaches point B is zero, all of its mechanical energy is stored as kinetic energy.
∑E = KE = KE(tran) + KE (rot) = 7/10 * Mv^2 = 437.6064J
7/10 * 4.8 * v^2 = 437.6064J
v^2 = 130.24
v = sqrt(130.24) = 11.4 m/s.
Therefore, the velocity of the ball when it reaches point B is 11.4 m/s.
Does this answer your question?