Q19P - Fundamentals Of Physics

Question

What are the magnitudes of

 (a) the angular velocity, 

(b) the radial acceleration, and 

(c) the tangential acceleration of a spaceship taking a circular turn of radius $\mathbf{3220}\mathbf{ }\mathbf{\text{km}}$at a speed of $\mathbf{29}\mathbf{ }\mathbf{000}\raisebox{1ex}{${\displaystyle \mathbf{\text{km}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{h}$}\right.$?

Step-by-Step Solution

Answer

1. The magnitude of the angular velocity of the spaceship is,$2.5x{10}^{3}rad/s$ .
2. The magnitude of the radial acceleration of the spaceship is,$20.2\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$
3. The magnitude of the tangential acceleration of the spaceship is,$\raisebox{1ex}{${\displaystyle \text{0m}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle {\text{s}}^{\text{2}}}$}\right.$

Step 1: Listing the given quantities

The radius of the circular path$r=3220 \text{km}=3.220 \times {10}^6 \text{m}$

The speed of the spaceship$v=29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.$

 

Step 2: Understanding the kinematic equations

The spaceship is taking a turn along a circular path. So, we need to use the equations that relates the linear variables (v and a) with the corresponding angular variables ($\omega$ and$\alpha$ ). The two variables are related through the parameter r, the radius of the path.

Formula:

$v= r\omega$

$a_r= \frac{v^2}{r}$

$a_t= \alpha r$

Step 3: Conversion of km/hr to m/s

We convert the speed of the spaceship to the units of m/s 

$\begin{array}{rcl}v& =& 29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.\\ & =& 29000\raisebox{1ex}{${\displaystyle \text{km}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \text{hr}}$}\right.\times \frac{1000 \text{m}}{1 \text{km}}\times \frac{1 \text{hr}}{3600 \text{s}}\\ & =& 8.05 \times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

 

Step 4: (a) Calculations of angular velocity

The angular velocity and the linear velocity are related as 

$\omega = \frac{v}{r}$

Substitute all the value in the above equation.

$$\begin{gathered} \omega =\frac{{\displaystyle 8.05\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.}}{{\displaystyle 3.22\times {10}^{6}m}} \\ \omega =2.5\times {10}^{-3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The magnitude of the angular velocity of the spaceship is$2.5\times {10}^{-3}\raisebox{1ex}{${\displaystyle \text{rad}}$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Step 5: (b) Calculations of radial acceleration

The radial acceleration is calculated as

$a_r= \frac{v^2}{r}$

 

Substitute all the value in the above equation.

$$\begin{gathered} a_r=\frac{{\displaystyle {\left(8.05\times {10}^3\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)}^2}}{{\displaystyle 3.22\times {10}^6 \text{m}}} \\ {a}_r=20.2\raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right. \end{gathered}$$

The magnitude of the radial acceleration of the spaceship is $20.2\raisebox{1ex}{${\displaystyle m}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle s}$}\right.$.

 

Step 6: (c) Calculations of tangential acceleration

The tangential acceleration can be calculated as $a_t= \alpha r$ where $\alpha$ is the angular acceleration. This is related to the change in the magnitude of the linear velocity v. The given spaceship is taking a turn at constant linear velocity. Hence, the magnitude of the velocity is constant. So, the angular acceleration  is zero. Hence, the tangential component of the linear acceleration is also zero.

$$\begin{gathered} a_t= \alpha r \\ {a}_t=0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

The magnitude of the tangential acceleration of the spaceship is,$0\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ .