A hollow sphere of radius 0.15 m, with rotational inertia I = 0.054 kg m2 about

Question

A hollow sphere of radius 0.15 m, with rotational inertia I = 0.054 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 38° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 100 J.
(a) How much of this initial kinetic energy is rotational?
J

(b) What is the speed of the center of mass of the sphere at the initial position?
m/s
Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?
J

(d) What is the speed of its center of mass now?
m/s

Explanation / Answer

         I = 0.054 kg .m2    =380 initial kinetic energy is K = 100 J
a)
           The initial kinetic energy is K = (1/2)I2 +(1/2)mU2                                               = (1/2)I( U/R)2 + (1/2)mU2                                                       = (1/2)(2/3)mR2* U2 / R2 +(1/2)mU2                                                       = (5/6)mU2                                 ==>              mU2 = 6(100 / 5)                                                          = 120J
        Then the rotationalkinetic energy is (1/2)I2 = K -(1/2)mU2                                                                               =100 J - ( 120/2)                                                                               =40 J      

b)
        The moment ofinertia is I = (2/3)mR2                                                          m = 3I / 2R2                                                            =3 ( 0.054 kg m2) / 2 ( 0.15 m)2
                                                          =3.6 kg
       Therefore
    mU2 = 120 J                     U = ( 120/ m)1/2                         = 5.77 m/s
    
   (c)   From the conservation ofenergy we have
                                    K1 + P.E1 = K2 + P.E2                                                 K2 = K1 + P.E1 - P.E2                                                       = 100J +(0 - m g h)                                                         = 100J - mg (1.0m(sin38))                                                         = 78.27 J       
(d) The total kinetic energy is K2 = (5/6)mV2                  ==> V = [ 6(78.27 J) / 5 ( 3.6 kg) ]1/2                               = 5.10 m/s    =380 initial kinetic energy is K = 100 J            The initial kinetic energy is K = (1/2)I2 +(1/2)mU2                                               = (1/2)I( U/R)2 + (1/2)mU2                                                       = (1/2)(2/3)mR2* U2 / R2 +(1/2)mU2                                    ==>              mU2 = 6(100 / 5)                                                          = 120J
        Then the rotationalkinetic energy is (1/2)I2 = K -(1/2)mU2                                                                               =100 J - ( 120/2)                                                                               =40 J      

b)
        The moment ofinertia is I = (2/3)mR2                                                          m = 3I / 2R2                                                            =3 ( 0.054 kg m2) / 2 ( 0.15 m)2
                                                          =3.6 kg
       Therefore
    mU2 = 120 J                     U = ( 120/ m)1/2                         = 5.77 m/s
    
   (c)   From the conservation ofenergy we have
                                    K1 + P.E1 = K2 + P.E2                                                 K2 = K1 + P.E1 - P.E2                                                       = 100J +(0 - m g h)                                                         = 100J - mg (1.0m(sin38))                                                         = 78.27 J       
(d) The total kinetic energy is K2 = (5/6)mV2                  ==> V = [ 6(78.27 J) / 5 ( 3.6 kg) ]1/2                               = 5.10 m/s                         = 5.77 m/s
    
   (c)   From the conservation ofenergy we have
                                    K1 + P.E1 = K2 + P.E2                                                 K2 = K1 + P.E1 - P.E2                                                       = 100J +(0 - m g h)                                                         = 100J - mg (1.0m(sin38))                                                         = 78.27 J       
(d) The total kinetic energy is K2 = (5/6)mV2                  ==> V = [ 6(78.27 J) / 5 ( 3.6 kg) ]1/2                               = 5.10 m/s                  ==> V = [ 6(78.27 J) / 5 ( 3.6 kg) ]1/2                               = 5.10 m/s                               = 5.10 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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