Question

I don't understand how to find 6b and all of 7.

6. A hollow sphere with mass = 0.65 kg and radius = 0.13 m is initially at rest on a 20° incline and rolls down the incline without slipping. The initial height of the disk (H) a. At the top of the incline, just before the disk begins to roll, what is the total mechanical energy of the disk? Emech=PEtop=6.37) b. Determine the velocity of the disk at H= 1.0 m the bottom of the incline. 200 PEtop=KEtran+KErot=6.37) KEtran=(1/2)mv2 and KErot=(1/2)Io2=(1/3)mv2 → (5/6) mv2=6.37) → Vbottom=3.43m/s c. Calculate the translational and rotational kinetic energy, respectively, of the ball when it reaches the bottom of the incline. KEtran=3.82) and KErot=2.54) 7. A figure skater with an initial moment of inertia of 75.0 kg m spins at a rotational speed of 180 rpm. Assume there is no friction acting on the skater. a. What is the angular momentum of the skater (in SI units)? 0=18.8rad/s → L=1410kg m/s b. What is the rotational kinetic energy of the skater? Nation KErot=1.33x104) XBI C. If the skater decreases his rotational inertia to 45.5 kg m2, determine the skater's new rotational speed? 0=31.1rad/s d. Calculate the skater's rotational kinetic energy in (c)? KErot=2.17x104)

Answer 1

6 Now as disk moves down it will Ratate and translate (Role). so it will have both Rotational. Kinetic Energy 8 translational L.E. Rike - 's I wv2 w=Ž ok D - R.K.EE - Total Energy = !mv't mu? - smr? - This energy comes from RE § mr? 6.37 606:37 560.65 3.43 | mis

wa 180 rpm wo 180 x 25 wo on :- 18.84 had I see E: Iw 2= 75x 18.84 = [1413 kg m3/s -€ sa - ), то? =\x75 118.845? 13310.46 -- |1:3310 x 10"] Angular momenten Iw, = Iz W2 11.33 10 x 10 is is conserved : رده وقال Iz 75* 18.84 45.5 : 310 sua

છે) K. Eget= - ૧૦ 1 - ૫૬ 50 (3) ૦૬ ૧) - 24૫૦. ૨ss | - .૧૫ ]