Question

Can you please solve the problem in detail because there are some concepts that I don’t understand

MOHAMED ALDEGN Sin 25 0.42 Cos 25 0.91 Tan 250.47 p 10 m's 3s 1994M2. Alarge phere rolls without slipping across a Note: Diagram not drawn to scale. translational speed of 10 meters per second, a mass m of 25 kilograms, and a radms the sphere about its ceater of mass is I-2ms. The sphere approache shown above and rolls up the incline without slipping and a radaus r of 02 meter. The moment of inertia of a 25° incline of beight 3 meters as a Calculate the total kinetic energy of the spbere as it rolls along the c. Neglecting air resistance, caleulate the boxizontal distance fiom the d. Suppose, instead, that the sphere horizontal surface. b i Calculate the nuagnitude of the spbere's velocity jast as it leaves t the incline as it leaves the top of the incline. the top of the incline point where the sphere leaves the incline to ii. Specify the direction of the sphere's velocity just the point where the sphere strikes the level surface. were to roll toward the incline as stated above, but the incline were State whether the speed of the sphere just as it leaves the top o tue greater than the speed caleulated in b. Explain brieftly. the incline would be less than, equal to, or briely. 1994V2 ii. It is directed parallel to the 25 0m-3m+(7.56 m's)Xsin 25%+(9.8 m/s)P which gives t-1.16 s (positive root) x-vt-(756 m'scos 251.16 s)- 793 m The speed would be less than in b. The gain in potential energy is entirely at the espense of the translational kinetic energy as there is no torque to slow the rotation d.

Answer 1

a.) As we know KE = translational KE + Rotational KE = 1/2 mv^2 + 1/2 I omega^2 = 1/2 mv^2 + 1/2 times 2/5 mR^2 times (v/R)^2 = 1/2 mv^2 + 1/5 mv^2 = 7/10 mv^2 = 7/10 times 25 times (10)^2 = 1750 J Ans b.) (i) At top it's velocity will be KE_horizontal = KE_top + PE_top 7/10 mv^2 = 7/10 mv_1^2 + mgh 7/10 times (10)^2 = 7/10 times v_1^2 + 9.8 times 3 v_1 = 7.62 m/s Ans (ii) As it will leave top incline than it's direction of motion is along 25 degrees from +ve x axis c.) Time for which ball remain in air T = 2v_1 sin theta/g + squareroot 2h/g = 2 times 7.62 times sin 25/9.81 + squareroot 2 times 3/9.8 = 0.66 + 0.78 = 1.44 s Hence distance traveled d = v_1 cos theta T = 7.62 times cos(25) times 1.44 = 9.94 m Ans