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P Pearson Sign In Pearson Sign In C Get Homework Help With Chex jay maleries tatton lubtock - GX 7 - PHIVS 1408, section 090 X + - X E → C webassign.net/web/Student/Assignment-Responses/last?dep=22467261 For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Assignment Scoring Your last submission is used for your score. 8. + -13 points SerPSE 10 10.9.OP.031. My Notes Ask Your Teacher A solid rubber ball is released from the top of an Incline of helght h that makes an angle with the horizontal. (a) If the ball rolls without slipping, what is its speed when It reaches the bottom of the Incline? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.) (b) If the ball Instead slides without rolling and without friction, what is its speed when it reaches the bottom of the Incline? (Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.) V = (c) Compare the time intervals required to reach the bottom in cases (a) and (b). Frolling sliding Need Help? Read It Submit Answer Practice Another Version View Previous Question Question 8 of 16 View Next Question 9:13 AM E Type here to search 10/16/2019 21

Answer 1

L EILA say, the mass of ball = m. radius of the ball =8. Initially the ball has only potential energy at the top of the inclined plane that is , P.E. = mgh.. But when it starts moving (rolling with -out slipping), it achieves a rotational Kinetic energy as well as a linear an translational kinetic energy. so the P.E. (ugh) converts into these kinetic energy when it reaches the bottern of the plane. If, & be the velocity at bottom, then total kinetic energy, K.E. = (mv)+ ( I wn) (1) moment of inertia of sphere about Its centre of mass and has a value of ( 2 mp). (w) angular velocity of the ball,

LLLLLLLLLL Now, after reaching at bottom , change in P.E. = gain in koe. I on, mgh = 12 mw " + L I on - Imv + 2 [ mn] Lage = 1 mult I muo = mun ( + 2 + 1) = 7 min on = √ 1 f g h 6 velocity at bottom for rolling without stapping.

ULUM 6 For sliding without rolling, there is only translatial kinetic energy Ct mvong and no rotational kinetic energy is present. So, for previous problem, if the ball slides only without rolling, then the initial potential energy P. E. (mgh) totally convents into translational KOE. Et mur), where is the velocity of the ball at bottam. So, In this case, trio migh = Imun or, (v = V2 gh] I LS-> INQ for both cases, the ball have to cover TB a linear distance (AB) to reach the bottom point. AB = h.cosec o hino ☺ for case , rolding without slipping, the linear accoaration of the ball, gsino Impw wwwWMAN

or Apolling=it Ihr g sino por sino here, the Ismoment of inentia of the ball = 2/3 mar. So, a polling = g 1 2 = 5 g sino If time taken to reach the bottom point is trolling, then AB = ustpolling & I. andling trolling initial speed, ere, fino a 5 asime x tisuing - or 14h - trolling 5 g sino

for care 6 , sliding with out rolling, the linear acceleration is only yg sino). arq a sliding = g sino so, il time taken to reach the bottom point in this case is tsliding & then AB = Un tsliding tot a sleding itsleding as there = { Casino). tšliding 74 7 initially on tsleding sino 127 So, the ratio of (trolling : tsliding) = sino S 14 h stror 2h trolling - 7 = 57:55 Itsliding WWWWWWWWWWWW