Figure 10-33 gives angular speed versus time for a thin rod that rotates around one end. The scale on the v axis is set by (a) What is the magnitude of the rod’s angular acceleration? (b) At t = 4.0s , the rod has a rotational kinetic energy of 1.60J. What is its kinetic energy at t = 0 ?
By using the concept of Angular acceleration, Rotational kinetic energy and moment of inertia we can find angular acceleration and rotational kinetic energy at t = 0 s. The formulas used for the concept are given below.
From the graph,
Angular acceleration is the rate of change of angular velocity.
Using all the values,
Hence the value of angular acceleration is, 1.5rad/s2.
Now to solve part b, we will calculate the inertia of the thin rod using rotational kinetic energy
Rearranging rotational kinetic energy formula for inertia I
Rotational inertia of the thin rod (I) is 0.2kh.m2 , we will use this to find rotational kinetic energy at,
Hence the kinetic energy of the thin rod at t = 0 is 0.4J.
A flywheel with a diameter of is rotating at an angular speed of .
(a) What is the angular speed of the flywheel in radians per second?
(b) What is the linear speed of a point on the rim of the flywheel?
(c) What constant angular acceleration (in revolutions per minute-squared) will increase the wheel’s angular speed to in ?
(d) How many revolutions does the wheel make during that ?
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length . At the instant it makes an angle of with the vertical as it falls, what are
(a) the radial acceleration of the top, and
(b) the tangential acceleration of the top. (Hint: Use energy considerations, not a torque.)
(c) At what angle is the tangential acceleration equal to g?
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