Question

One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of radius R and frequency f. The tension in the string is Fs . If both the radius and frequency are doubled, the tension is what

Answer 1

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration ($a_{r}$), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

$a_{r} = 4\pi^{2}\cdot f^{2}\cdot R$ (1)

Where:

$f$ - Frequency, measured in hertz.

$R$ - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension ($T$), measured in newtons, through the string, then we derive the following model:

$\Sigma F = T = m\cdot a_{r}$ (2)

Where $m$ is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

$T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R$ (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

$\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right)$ (4)

$\frac{T_{2}}{T_{1}} = 4\cdot 2$

$\frac{T_{2}}{T_{1}} = 8$

If both the radius and frequency are doubled, then the tension is increased 8 times.