An object of mass 0.5kg is rotated in horizontal circle by a string of length 1m,the maximum tension in the string before it breaks is 50N.What is the greatest number of revolution per second of the object?
tension=mw^2/r
50=.5*w^2/1
w^2=100
w=10 rad/sec=10/2PI rev/sec
why is length of the string equal to the radius pls explain with example
isnt the formula mv^2/r?
To find the maximum number of revolutions per second of the object, we first need to find the maximum velocity of the object as it rotates in the horizontal circle. We can use the concept of centripetal force to determine the maximum velocity.
The centripetal force acting on an object is given by the equation:
Fc = (m * v^2) / r
Where Fc is the centripetal force, m is the mass of the object, v is the velocity, and r is the radius (in this case, the length of the string).
In this case, the maximum tension in the string before it breaks is given to be 50N. This tension is equivalent to the centripetal force acting on the object, since the string is responsible for providing the necessary centripetal force to keep the object in circular motion.
Therefore, we have:
Fc = 50N
m = 0.5kg
r = 1m
Substituting these values into the centripetal force equation, we can solve for the maximum velocity:
50N = (0.5kg * v^2) / 1m
Rearranging the equation, we get:
v^2 = (50N * 1m) / 0.5kg
v^2 = 100 Nm/kg
Taking the square root of both sides, we find:
v = √(100 Nm/kg)
v ≈ 10 m/s
Now, the number of revolutions per second can be calculated using the equation:
f = v / (2πr)
where f is the frequency (revolutions per second) and r is the radius.
Substituting the values:
v = 10 m/s
r = 1 m
f = 10 m/s / (2π * 1 m)
f ≈ 1.59 revolutions per second
Therefore, the greatest number of revolutions per second of the object is approximately 1.59.
tension=mw^2/r
50=.5*w^2/1
w^2=100
w=10 rad/sec=10/2PI rev/sec