Asolid sphere with a diameter of 0.14 m is released from rest; it then rolls without slipping down a ramp, dropping through a vertical height of h1 = 0.70 m. the ball leaves the bottom of the ramp, which is h2 = 1.08 m above the floor, moving horizontally.

👤julianna 9✉ Mail âš™ Help Science & Mathematics Physics AnDrew A solid sphere with a diameter of 0.19 m is released from rest; it then rolls without slipping down a ramp, dr A solid sphere with a diameter of 0.19 m is released from rest; it then rolls without slipping down a ramp, dropping through a vertical height of h1 = 0.51 m. The ball leaves the bottom of the ramp, which is h2 = 1.35 m above the floor, moving horizontally. (a) Through what horizontal distance d does the ball... show more 1 following 3 answers Report Abuse Answers Relevance MikeyZ Best Wow, this is sorta tough. Okay, I will start by guessing that "horizontal distance" is relative to the bottom of the ramp, not the top (since we don't know the angle of the ramp anyway, it's moot). You solve this problem by evaluating the kinematics of the sphere at the bottom of the ramp. The easiest way to solve this is to invoke conservation of energy. At the top of the ramp, it's all potential energy, U = mgh, where m is the sphere mass, g is gravity constant, and h is the vertical displacement of the ball from top to bottom, which you can show is just the height of the ramp. At the bottom of the ramp, it's all kinetic energy (note: to be pedantic, yes, there is more "potential energy" in the transition from the top of the ramp to the floor, but since I'm referencing my initial potential energy relative to the base of the ramp and not the floor, that's where I'm calling my "zero-potential-energy" point. You can reference it from wherever you want, the floor, the ramp, or any arbitrary point, and the answer will always be the same). To solve kinetic energy, you have two terms, rotational and translational, and because the ball rolls without slipping the velocity is tied to the angular velocity by the radius of the ball: K = (1/2) mv² + (1/2) Iω², where m is the ball mass, I is the ball's moment of inertia (2/5)mr², and ω is the angular velocity of the ball. Because the ball rolls without slipping, it is easy to see that v=ωr, or r=v/ω. Then, K = (1/2)mv² + (1/2)(2/5)mr²ω² = (1/2)mv² + (1/5)mv² = (7/10)mv² Setting potential at the top equal to kinetic at the bottom, mgh=(7/10)mv² v=âš{(10/7)(gh)} = [(10/7)(9.8)(0.51)]^(1/2) = 2.672m/s Now to answer the rest of the question, we need to make an assumption. Since we weren't told what the angle of the ramp is, we don't know what the initial velocity (speed and angle) of the falling ball is. I think it is probably reasonable to assume that, because the ball is so high off the ground (h2), that the prevailing speed when it hits the floor would be basically the same as if the ball fell from rest. That is to say, launching a sphere from a steep ramp would have a decidedly different outcome than launching a sphere from a gradual ramp, even if the heights were the same. So I'm going to assume that the ramp is very gradual. Then we can conclude that the initial vertical speed of the sphere upon release is negligble, and d = vt Get t from y = (1/2)gt²; t=âš(2y/g) = âš(2*1.35/9.8) = .525s (a) d=.525*2.672 = 1.40m The total subtended angle of rotation is just the product of angular velocity times time: θ = ωt = vt/r = d/r = 1.40m/0.19m = 7.38 radians The total number of revolutions is then (b) n=θ/(2Ď€) = 1.17 revs (c) If the ramp were frictionless, there would be no rolling motion, and all the kinetic energy would be translational. That means, for constant potential energy, the speed of the sphere would be higher when it left the ramp. That means the distance d would increase for the same fall time.