a 10 kg mass is attached to one end of a string with a lenght of 5m and whirled in a horizontal circle overheadd at a rate of 1 revolution per second what is tension in the sring
It is impossible of course to make that circle exactly horizontal since tan theta = g/Ac
so we must assume that g is small or Ac is much much bigger than g. Let's see:
Ac = v^2/R = w^2 r
where w is omega, radians/second
omega = 2 pi radians/second
so
Ac = (2 pi)^2 (5) = 197 m/s^2
yes, that is much bigger than g so circle is nearly horizontal
F = m Ac = 10*197
= 1970 Newtons in the string
To determine the tension in the string, we first need to understand the forces acting on the mass as it moves in a horizontal circle.
In circular motion, there are two primary forces at play: the centripetal force (toward the center of the circle) and the tension force in the string.
The centripetal force (F_c) is given by the equation:
F_c = (m * v^2) / r
Where:
m = mass of the object (10 kg in this case)
v = linear velocity of the object
r = radius of the circular path (equal to the length of the string, 5m in this case)
In this scenario, the object completes one revolution (a full circle) per second.
To find the linear velocity (v), we can use the formula:
v = 2 * π * r / T
Where:
T = time taken to complete one revolution (1 second in this case)
Substituting the values, we have:
v = 2 * π * 5 / 1
So the linear velocity (v) is 10π m/s.
Now we can plug this value into the centripetal force equation:
F_c = (10 * (10π)^2) / 5
Simplifying this expression, we get:
F_c = 200π^2 N (Newtons)
Since there are no other vertical forces acting on the object, the tension in the string (T) must be equal to the centripetal force (F_c), as:
T = F_c = 200π^2 N
Therefore, the tension in the string is approximately 200π^2 N.