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Fundamentals Of Physics
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Short Answer

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 ω=6.0rad/s (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 ?

  1. Magnitude of the rod’s angular acceleration is, α=1.5rad/s2
  2. The kinetic energy when t = 0 is K = 0.4J.
See the step by step solution

Step by Step Solution

Step 1: Understanding the given information

  1. Angular velocity of the wheel is, ω2=6.0rad/s
  2. Kinetic energy of the wheel, at t =4 s, K = 1.60 J

Step 2: Concept and Formula used for the given question

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.

  1. Angular acceleration, =dωdt
  2. Rotational Kinetic energy, K=12lω2

Step 3: (a) Calculation for the magnitude of the rod’s angular acceleration

From the graph,

At,

ti=0s, ωi=-2rad/s

And,

tf=4s, ωf=4rad/s

Angular acceleration is the rate of change of angular velocity.

α=dωdt=ωf-ωitf=ti

Using all the values,

α=4rad/s--2rad/s4s-0=64rad/s2=1.5rad/s2

Hence the value of angular acceleration is, 1.5rad/s2.

Step 3: (b) Calculation for the kinetic energy at t=0

Now to solve part b, we will calculate the inertia of the thin rod using rotational kinetic energy K=1.60J at t=4s

Rearranging rotational kinetic energy formula for inertia I

K=12lω2l=2Kω2

We have

at t=4s, K=1.60J, ω=4rad/sl=2×1.60J4rad/s2=0.2kg.m2

Rotational inertia of the thin rod (I) is 0.2kh.m2 , we will use this to find rotational kinetic energy at,

t=0s, ω=-2rad/sK=12lω2=12×0.2kg.m2×-2rad/s2=0.4J

Hence the kinetic energy of the thin rod at t = 0 is 0.4J.

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