Question

63 A model airplane of mass 0.750kgflies with a speed of 35.0 m/sin a horizontal circle at the end of a 60.0mlong control wire as shown in above figure a: The forces exerted on the airplane are shown in above figure b: the tension in the control wire, the gravitational force, and aerodynamic lift that acts at $\mathbf{\theta }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$inward from the vertical. Compute the tension in the wire, assuming it makes a constant angle of $\mathbf{\theta }\mathbf{=}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$with the horizontal.

Short answer

The tension in the wire is 12.8N.

Step-by-step solution

Stating given data

Mass of the model airplane m=0.750kg

Speed of the plane moving in horizontal circle v= 35.0 m/s

Length of the control wire = 60.0m

Defining centripetal acceleration

The force required to move the particle in circular motion is known as centripetal force.

The expression for the centripetal force is given by

$\mathbf{F}\mathbf{=}\mathbf{m}\frac{{\mathbf{v}}^{\mathbf{2}}}{\mathbf{R}}$

Here, m is the mass of the object, v is the speed and R is the radius, F is the centrifugal force. Centrifugal force is also known by the name of centripetal force.

Calculating the tension in the wire

The plane's acceleration is toward the center of the circle of motion, so it is horizontal. The radius of the horizontal circular path is,

$$\begin{gathered} \mathrm{r}=(60.0\mathrm{m})\cos 20.0^0 \\ =56.4\mathrm{m} \end{gathered}$$

The centripetal acceleration is given by,

${\mathrm{a}}_{\mathrm{c}}=\frac{{\mathrm{v}}^2}{\mathrm{r}}$

Here ${\mathrm{a}}_{\mathrm{c}}$is the centripetal acceleration, v is the speed and r is the radius of the circular path.

Substitute 35m/s for v, 56.4m for r in the above equation.

$\begin{array}{c}{a}_c=\frac{{(35)}^2}{56.4}\\ =21.7m/s^2\end{array}$

We can also calculate the weight of the airplane: