Category Archives: Physics

Acceleration of a Revolving Ball – Uniform Circular Motion Example

A 150-g ball at the end of a string is revolving uniformly in a horizontal circle of radius 0.600 m, as in the Figure 1 below. The ball makes 2.00 revolutions in a second. What is its centripetal acceleration?

Figure 1: A small object moving in a circle, showing how the velocity changes. At each point, the instantaneous velocity is in a direction tangent to the circular path.

Solution:

The linear velocity of the ball can be computed by dividing the total arc length traveled by the total time of travel. That is, the ball traveled 2 revolutions (twice the circumference of the circle) for 1 second. Thus,

\begin{align*}
\text{v} &= \frac{2\cdot2 \pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi \text{r}}{\text{t}} \\
\\
& = \frac{4\pi\left( 0.600\ \text{m} \right)}{1 \ \text{s}} \\
\\
& = 7.54 \ \text{m/s}
\end{align*}

Since the linear velocity has already been computed, we can now compute for the centripetal acceleration, ac.

\begin{align*}
\text{a}_\text{c} & = \frac{\text{v}^{2}}{\text{r}} \\
\\
& = \frac{\left( 7.54\ \text{m/s} \right)^{2}}{0.600\ \text{m}}\\
\\
& =94.8 \ \text{m/s}^{2}
\end{align*}

Physics Problem: Fast-Plant

Bamboo grows very fast, for a plant. A particular bamboo might grow 90.4 cm in a single day. How many micrometers per second does this rate correspond to?

Solution:

\begin{aligned}
\text{rate} & = \frac{\text{length}}{\text{time}} \\
\\
&=\frac{90.4\  \text{cm}}{1 \ \text{day}}\\
\\
&=\frac{90.4\ \bcancel{\text{cm}}}{\bcancel{\text{day}}} \times\frac{10,000\ \mu m}{1\ \bcancel{\text{cm}}}\times \frac{1\ \bcancel{\text{day}}}{86400\ s}\\
\\
& = 10.5\ \mu m/s
\end{aligned}

College Physics 2.51 – Time of the hiker to move out from a falling rock

Standing at the base of one of the cliffs of Mt. Arapiles in Victoria, Australia, a hiker hears a rock break loose from a height of 105 m. He can’t see the rock right away but then does, 1.50 s later. (a) How far above the hiker is the rock when he can see it? (b) How much time does he have to move before the rock hits his head?

Solution:

Part A

We know that the initial height, y_0 of the rock is 105 meters, and the initial velocity, v_0 is zero. We shall solve for the distance traveled by the rock for 1.5 seconds from the initial position first to find the height at detection.

The change in height is

\displaystyle \begin{aligned}
\Delta \text{y}&=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2 \\
&=\left( 0  \right)\left( 1.50 \ \text{s} \right)+\frac{1}{2}\left( 9.81\ \text{m/s}^{2} \right)\left( 1.50\ \text{s} \right)^{2}\\
&=0+11.036\ \text{m} \\
&=11.04 \ \text{m} 
\end{aligned}

So, the rock falls about 11.04 m from the initial height for 1.50 seconds. Therefore, the height of the rock above his head at this point is

\displaystyle \begin{aligned}
\text{y}&=\text{y}_{0}-\Delta \text{y} \\
&=105\ \text{m}-11.04\ \text{m} \\
&=93.96 \ \text{m}
\end{aligned}

Part B

We shall solve for the total time of travel, that is, from the initial position to his head. Then we shall subtract 1.50 s from that to solve for the unknown time of moving out. The total time of travel is

\begin{aligned}
\text{y} & =\frac{1}{2}\text{at}^{2} \\
&\text {Solving for t, we have}\\
\text{t}&=\sqrt{\frac{\text{2y}}{\text{a}}} \\
&=\sqrt{\frac{2\left( 105\ \text{m} \right)}{9.81 \ \text{m/s}^{2}}} \\
&=4.63 \ \text{s}

\end{aligned}

Therefore, to move out the hiker has about

\begin{aligned}
\text{t}&=4.63 \ \text{s}-1.50\ \text{s}\\
&=3.13\ \text{s}
\end{aligned}

College Physics 2.50 – Motion of a Jumping Kangaroo

A kangaroo can jump over an object 2.50 m high. (a) Calculate its vertical speed when it leaves the ground. (b) How long is it in the air?

Part A

The motion of the kangaroo is under free-fall. We are looking for the initial velocity, and we know that the velocity in the highest position is zero.

From 

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay},\\
\end{aligned}

we have

\begin{aligned}
\text{v}^2 &=\left (\text{v}_0 \right )^2+2\text{ay}\\
\text{v}^2-2\text{ay} &= \left ( \text{v}_0\right)^2\\
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}}
\end{aligned}

Substituting the known values, 

\begin{aligned}
\text{v}_0&=\sqrt{\text{v}^2-2\text{ay}} \\
\text{v}_0&=\sqrt{0^2-2\left(-9.81 \text{m/s}^2\right)\left(2.50 \text{m}\right)}\\
\text{v}_0&= {\color{green}7.00 \  \text{m/s}}
\end{aligned}

Therefore, the vertical speed of the kangaroo when it leaves the ground is 7.00 m/s.

Part B

Since the motion of the kangaroo has uniform acceleration, we can use the formula

\text{y}=\text{v}_o\text{t}+\frac{1}{2}\text{a}\text{t}^2

The initial and final position of the kangaroo will be the same, so y is equal to zero. The initial velocity is 7.00 m/s, and the acceleration is -9.81 m/s2.

\begin{aligned}
\text{y} & =\text{v}_0\text{t}+\frac{1}{2}\text{a}\text{t}^2\\
0 & = \left( 7.00\ \text{m/s} \right)\text{t}+\frac{1}{2}\left( -9.81\ \text{m/s}^{2} \right)\text{t}^2\\
0 & =7\text{t}-4.905\text{t}^{2}\\
7\text{t}-4.905\text{t}^{2}&=0 \\
\text{t}\left( 7-4.905\text{t} \right) & =0 \\
\text{t}=0 \qquad &\text{or} \qquad 7-4.905\text{t}=0 \\

\end{aligned}

Discard the time 0 since this refers to the beginning of motion. Therefore, we have

\begin{aligned}
7-4.905\text{t} &=0 \\
4.905\text{t} & = 7 \\
\text{t} & =\frac{7}{4.905} \\
 \text{t}&={\color{green}1.43 \  \text{s}} 
\end{aligned}

The kangaroo is about 1.43 seconds long in the air.

Solution Guides to College Physics by Openstax Chapter 13 Banner

Chapter 13: Temperature, Kinetic Theory, and the Gas Laws

Temperature

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Thermal Expansion of Solids and Liquids

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The Ideal Gas Law

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Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature

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Humidity, Evaporation, and Boiling

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Chapter 12: Fluid Dynamics and Its Biological and Medical Applications

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Bernoulli’s Equation

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The Most General Applications of Bernoulli’s Equation

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Viscosity and Laminar Flow; Poiseuille’s Law

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The Onset of Turbulence

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Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes

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Chapter 11: Fluid Statics

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Pressure

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Pascal’s Principle

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Archimedes’ Principle

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Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

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Solution Guides to College Physics by Openstax Chapter 10 Banner

Chapter 10: Rotational Motion and Angular Momentum

Angular Acceleration

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Kinematics of Rotational Motion

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Dynamics of Rotational Motion: Rotational Inertia

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Rotational Kinetic Energy: Work and Energy Revisited

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Angular Momentum and Its Conservation

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Collisions of Extended Bodies in Two Dimensions

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Gyroscopic Effects: Vector Aspects of Angular Momentum

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Solution Guides to College Physics by Openstax Chapter 9 Banner

Chapter 9: Statics and Torque

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The Second Condition for Equilibrium

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Stability

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Applications of Statics, Including Problem-Solving Strategies

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Simple Machines

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Forces and Torques in Muscles and Joints

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Solution Guides to College Physics by Openstax Chapter 8 Banner

Chapter 8: Linear Momentum and Collisions

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Linear Momentum and Force

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Impulse

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Conservation of Momentum

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Elastic Collisions in One Dimension

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Inelastic Collisions in One Dimension

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Collisions of Point Masses in Two Dimensions

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Introduction to Rocket Propulsion

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