Answer:
The speed of the cylinder has decreased
Explanation:
Given data:
Distance between the center of the turn table and the cylinder = R
Distance between the center of the turn table and the cylinder when it is moved to a new point, R' = R/2
the tangential velocity in circular motion is given as:
[tex]V=\frac{2\pi R}{T}[/tex]
where T is the time
now for R = R' = R/2
the new velocity V' comes as:
[tex]V'=\frac{2\pi \frac{R}{2}}{T}[/tex]
or
[tex]V'=\frac{\pi R}{T}[/tex]
thus the velocity becomes half or we can say the velocity decreases
Now, the acceleration (a) in circular motion is given as:
a = V²/R
now for
R = R' = R/2
the velocity V' = V/2
thus,
new acceleration a' comes as:
[tex]a = \frac{(\frac{V}{2})^2}{R/2}[/tex]
or
[tex]a'=\frac{V^2}{2R}[/tex]
thus,
the acceleration decreases by 2 times
The statement that accurately describes the motion of the cylinder at the new location is : ( 1 ) and ( 3 )
Given data :
Distance between cylinder and center of the turn table = R
Distance between cylinder and turntable when moved = R/2
where ;
Tangential Velocity/speed = [tex]\frac{2\pi R}{T}[/tex]
R = distance between objects
T = time
∴ The initial Tangential velocity of the cylinder at rest ( V ) = [tex]\frac{2\pi R}{T}[/tex] ----- ( 1 )
while the
Tangential velocity/speed of the cylinder when it moves ( V' ) = [tex]\frac{2\pi R/2}{T}[/tex]
= [tex]\frac{\pi R}{T}[/tex] ----- ( 2 )
Therefore comparing equations ( 1 ) and ( 2 ) the speed of the cylinder has decreased as the cylinder moves from R to R/2 .
Also Given that Tangential acceleration = v² / R .
Since velocity decreases after the cylinder moves, the magnitude of the acceleration of the cylinder will decrease as well.
Hence we can conclude that the acceleration and speed of the cylinder has decreased with the movement of the cylinder.
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