Physics, 19.11.2020 19:10, twistedhyperboles

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

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answered: lilybear1700

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

From kinematic formula:

$\bold {a_r = 4\pi r^2 \times f^2 \times R^2}$ .............................. (1)

Where:

f - Frequency,

R - Radius of rotation,

From Second Newton's Law,

$\bold {\sum F = T = m\times a_r }$ ......................... (2)

Where

m is the mass of the object,

From equation 1 and 2,

$\bold {T = m\times 4\pi r^2 \times f^2 \times R^2}\\$ .................... (3)

From equation, 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:

$\bold {\dfrac {T_2}{T_1} = (\dfrac {F_2}{F_1})^2 \times \dfrac {R_2}{R_1} = 4 \times 2 = 8 }$

Therefore, If both the radius and frequency are doubled, then the tension will be increased 8 times.

To know more about frequency,

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answered: amiechap12

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:

- Frequency, measured in hertz.

- Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (), measured in newtons, through the string, then we derive the following model:

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

Where 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)