Tag Archives: kinematics of Rotational Motion

College Physics by Openstax Chapter 2 Problem 62

By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s2 at t=10 s .

Figure 2.63

Solution:

To solve for the slope of the curve at t=10 s, we need two points — 1 just before and 1 just after.

When t=0 st=0 \ \text{s}, v=166 m/sv=166 \ \text{m/s} and when t=20 st=20 \ \text{s}, v=230 m/sv=230 \ \text{m/s}. Therefore, the acceleration is

acceleration=slopeacceleration=ΔvΔta=v2−v1t2−t1a=230 m/s−166 m/s20 s−0 sa=3.2 m/s2\begin{align*} \text{acceleration} & =\text{slope} \\ \text{acceleration} & =\frac{\Delta v}{\Delta t} \\ \text{a} & = \frac{v_2-v_1}{t_2-t_1} \\ \text{a} & = \frac{230\ \text{m/s}-166\ \text{m/s}}{20\ \text{s}-0\ \text{s}} \\ \text{a} & =3.2\ \text{m/s}^2 \\ \end{align*}

Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.

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Solution Guides to College Physics by Openstax Chapter 10 Banner

Chapter 10: Rotational Motion and Angular Momentum

Angular Acceleration

Problem 1

Problem 2

Problem 3

Problem 4

Kinematics of Rotational Motion

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Dynamics of Rotational Motion: Rotational Inertia

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Rotational Kinetic Energy: Work and Energy Revisited

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Angular Momentum and Its Conservation

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Collisions of Extended Bodies in Two Dimensions

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Gyroscopic Effects: Vector Aspects of Angular Momentum

Problem 48

College Physics by Openstax Chapter 2 Problem 22

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105 m/s2 for 8.10×10-4 s . What is its muzzle velocity (that is, its final velocity)?

Solution:

We are given the following: a=6.20×105 m/s2a=6.20 \times 10^{5} \ \text{m/s}^2; Δt=8.10×10−4 s\Delta t=8.10 \times 10^{-4} \ \text{s}; and v0=0m/sv_0=0 \text{m/s}.

The muzzle velocity of the bullet is computed as follows:

vf=v0+atvf=0 m/s+(6.20×105 m/s2)(8.10×10−4 s)vf=502 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ v_f & = 0\:\text{m/s}+\left(6.20\times 10^5\text{ m/s}^2\right)\left(8.10\times 10^{-4}\:\text{s}\right) \\ v_f & =502\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Therefore, the muzzle velocity, or final velocity, is 502 m/s. 

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College Physics by Openstax Chapter 2 Problem 21

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Solution:

We are given the following values: a=−2.10×104 m/s2;t=1.85×10−3 s;vf=0 m/sa=-2.10 \times 10^4 \ \text{m/s}^2; t=1.85 \times 10^{-3} \ \text{s}; v_f=0 \ \text{m/s}.

The formula in solving for the initial velocity is

v0=vf−atv_0=v_f-at

Substitute the given values

v0=0 m/s−(−2.10×104 m/s2)(1.85×10−3 s)v0=38.85 m/s  (Answer)\begin{align*} v_0 & =0\:\text{m/s}-\left(-2.10\times 10^4\text{ m/s}^2\right)\left(1.85\times 10^{-3}\:\text{s}\right) \\ v_0 & =38.85\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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