Respuesta :
Answer:
The centripetal acceleration of the rock is decreases by a factor of 3.
Explanation:
Given that,
Radius of circle [tex]r= R[/tex]
Time period = T
Radius of new circle [tex]r'=\dfrac{R}{3}[/tex]
We need to calculate the velocity
Using formula of velocity
[tex]T=\dfrac{2\pi r}{v}[/tex]
The time period for new circle
[tex]T=\dfrac{2\pi r'}{v'}[/tex]
The time period remains same.
[tex]\dfrac{2\pi r}{v}=\dfrac{2\pi r}{v}[/tex]
Put the value in the equation
[tex]\dfrac{2\pi R}{v}=\dfrac{2\pi\dfrac{R}{3}}{v'}[/tex]
[tex]v=3v'[/tex]
The acceleration of the rock is
[tex]a=\dfrac{v^2}{R}[/tex]
We need to calculate the acceleration of the new rock
Using formula of acceleration
[tex]a'=\dfrac{v'^2}{r'}[/tex]
Put the value into the formula
[tex]a'=\dfrac{\dfrac{v^2}{9}}{\dfrac{R}{3}}[/tex]
[tex]a'=\dfrac{v^2}{3R}[/tex]
[tex]a'=\dfrac{1}{3}a[/tex]
Hence, The centripetal acceleration of the rock is decreases by a factor of 3.
Answer:
Explanation:
Let the mass of rock is m.
Initial radius = R
initial time = T
final radius = R / 3
The initial centripetal acceleration is given by
[tex]a=\frac{\omega ^{2}}{R}=\frac{4\pi ^{2}}{RT^{2}}[/tex] .... (1)
Now the new centripetal acceleration is given by
[tex]a'=\frac{\omega ^{2}}{\frac{R}{3}}=\frac{3\times 4\pi ^{2}}{RT^{2}}=3 a[/tex] from equation (1)
So, the new centripetal acceleration is 2 times the initial acceleration.
