Suppose a cylindrical platform with moment of inertia 0.005kg m2 is freely rotat

Question

Suppose a cylindrical platform with moment of inertia 0.005kg m2 is freely rotating at angular speed 10 rad/s. A disc initially, rotationally at rest with mass 0.6kg and radius 0.1 m is dropped onto the platform so that their centers coincide. The kinetic frictional coefficient between the faces of the platform and disk is 0.25. Eventually both rotate with a common speed. Calculate the final speed.
1.How much energy is lost in the collision?
Also, assuming that all the energy is lost to friction and that the torque due to friction is 1/2FkR, determine the relative slide angle of the disk over the platform.
Finally, how much time does it take for the disk to reach the final speed?

Explanation / Answer

Given Moment of inertia of the patform, Ipf = 0.005 kg m^2 Initial angular speed of the platform, i= 10 rad/s Mass of the disk, m = 0.6 kg Radius of the disk, r = 0.1 m Moment of inertia of the disk, Id= (1/2) m r2                                                    = (1/2) (0.6 kg) (0.1 m)^2                                                    = 0.003 kg.m^2                                                    = (1/2) (0.6 kg) (0.1 m)^2                                                    = 0.003 kg.m^2 Kinetic frictional coefficient between the faces of the platform and disk is, k= 0.25 Let the final velocity of the disk and platform be f By law of conservation of angular momentum Li = Lf Ipf i = (Ipf + Id) f (0.005 kg m^2) (10 rad/s) = (0.005 kg m^2 +0.003 kg m^2)f f = 6.25 rad /s is the final speed ------------------------------------------------------------------------------ 1. Initial energy of the system is,     Ki = (1/2) Ipf i2 = 0.5 (0.005 kg.m^2)(10 rad/s)^2                                 = 0.25 J Final energy of the system is,     Kf = (1/2)(Ipf + Id) f2 = 0.5 (0.005 kg.m^2+0.003 kg.m^2)(6.25 rad/s)^2                                           = 0.15 J Energy lost in the collision = 0.25 J - 0.15 J = 0.1 J ------------------------------------------------------------------------------- Torque due to friction is, = 1/2 Fk R Since = Fk R sin 1/2 Fk R=  Fk R sin sin = 1/2 = 30^0 Thus, the relative slide angle of the disk is 30^0 ---------------------------------------------------------------------------------- Time taken for the disk to reach the final speed is, f = i + t     ---- (1) (0.005 kg m^2) (10 rad/s) = (0.005 kg m^2 +0.003 kg m^2)f f = 6.25 rad /s is the final speed ------------------------------------------------------------------------------ 1. Initial energy of the system is,     Ki = (1/2) Ipf i2 = 0.5 (0.005 kg.m^2)(10 rad/s)^2                                 = 0.25 J Final energy of the system is,     Kf = (1/2)(Ipf + Id) f2 = 0.5 (0.005 kg.m^2+0.003 kg.m^2)(6.25 rad/s)^2                                           = 0.15 J Energy lost in the collision = 0.25 J - 0.15 J = 0.1 J ------------------------------------------------------------------------------- Torque due to friction is, = 1/2 Fk R Torque due to friction is, = 1/2 Fk R Since = Fk R sin 1/2 Fk R=  Fk R sin sin = 1/2 = 30^0 Thus, the relative slide angle of the disk is 30^0 ---------------------------------------------------------------------------------- Time taken for the disk to reach the final speed is, But angular accelertaion is calculated using the torque relation                      = I            1/2 Fk R = (0.003 kg m^2)            1/2 (km g ) R = (0.003 kg m^2)                   = 0.5(0.25)(0.6 kg)(9.8 m/s^2)(0.1 m) / (0.003 kg m^2)                      = 24.5 rad/s^2 Taking equation (1),                  6.25 rad/s = 0 + (24.5 rad/s^2)t                                 t = 0.25 s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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