The moment of inertia of a solid cylinder about its axis isgiven by 0.5 MR^2. If

Question

The moment of inertia of a solid cylinder about its axis isgiven by 0.5 MR^2. If the cylinder rolls without slipping, theratio of its rotational kinetic energy to its translational kineticenergy is

A wheel accelerates with a constant angular acceleration of 4.5rad/s^2. If the initial angular velocity is 1.0 rad/s, what is theangular velocity at t = 2.0 s?

A Ferris wheel rotating at 20 rad/s decelerates with a constantangular acceleration of -5.0 rad/s^2. How many revolutions does itrotate before coming to rest?

Explanation / Answer

Rotational kinetic energy K ' = ( 1/ 2) I w^ 2 where I = moment of inertia = 0.5 MR^2           w =angular speed = v / R So, K ' = ( 1/ 2) * 0.5 MR ^ 2 * ( V^ 2/ R^2 )            = 0.25 Mv^ 2 Translational kinetic energy K = ( 1/ 2) Mv^ 2 Therefore required ratio = K ' / K                                     = 0.5 (b).  angular acceleration = 4.5rad/s^2 initial angular velocity w = 1.0 rad/s the angular velocity at t = 2.0 s is w' = w +t                                                         = 1 + 9 = 10 rad / s (c). Initial angular speed w = 20 rad/s angular acceleration = -5.0rad/s^2 final angular speed w ' = 0 from the relation w' ^ 2- w^ 2= 2 angular displacement = [ w'^ 2- w^ 2] / 2                                  = 40 rad Therefore No.of revolutions does it rotate beforecoming to rest = / 2                                                                                                  = 6.366

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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