A 14 kg bowling ball with a radius of 11 cm starts from rest at the top of an in

Question

A 14 kg bowling ball with a radius of 11 cm starts from rest at the top of an incline 3.8 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume that the ball is a uniform solid sphere.)The acceleration of gravity is 9.81 m/s2. Answer in units of m/s A solid 218 N ball with a radius of 0.110 m rolls 7.8 m down a ramp that is inclined at 19 degree with the horizontal. If the ball starts from rest at the top of the ramp, what is the angular speed of the ball at the bottom of the ramp? The acceleration of gravity is 9.81 m/s2. Answer in units of rad/s

Explanation / Answer

The total energy of the ball on top of the incline is
Epot = mgh = 14*9.81*3.8 = 521.89 Nm

And the total energy changes into rotational and kinetic energy at the base of the incline:
521.89 Nm = 1/2 mv^2 + 1/2 I^2 and v = r
.................= 1/2 m(r^2^2) + 1/2 (2/5*mr^2)*^2
.................= 1/2 mr^2^2(1+2/5)
.................= 1/2*7/5*mr^2^2
.................= 7/10mv^2
v^2 = 521.89*10/(7m)
v^2 = 521.89*10/(7*14)=
v^2 = 53.254
v = 7.296 m/s = lateral velocity

Let me know how part a goes for u, if right can u rate it and then i'll do part b and send you the solution.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.