an object of mass 0.5 kg is whirled at the end of string 0.8 m long if the string make three revolution in 1.2 sec find the tension of the string
35.6N
I want the nice explanation how the answer came
To find the tension in the string, we can use the centripetal force formula:
F = (m * v^2) / r
Where:
F = Centripetal force (tension in the string)
m = Mass of the object = 0.5 kg
v = Velocity of the object
r = Radius of the circular path = 0.8 m
First, let's find the velocity of the object. We know that the string makes three revolutions in 1.2 seconds. The number of revolutions per second is given by the formula:
Revolutions per second = Total revolutions / Total time
In this case, the object makes 3 revolutions in 1.2 seconds. So:
Revolutions per second = 3 / 1.2 = 2.5 revolutions per second
The velocity of the object can be found by multiplying the angular speed (ω) with the radius (r):
v = ω * r
Since the object makes 2.5 revolutions per second, the angular speed (ω) can be found as:
Angular speed (ω) = 2π * Revolutions per second
ω = 2π * 2.5 = 15.71 rad/s (approximately)
Now, we can calculate the velocity of the object:
v = ω * r
v = 15.71 rad/s * 0.8 m
v = 12.57 m/s (approximately)
Finally, substitute the values of mass (m = 0.5 kg), velocity (v = 12.57 m/s), and radius (r = 0.8 m) into the centripetal force formula:
F = (m * v^2) / r
F = (0.5 kg * (12.57 m/s)^2) / 0.8 m
F = (0.5 kg * 157.84 m^2/s^2) / 0.8 m
F = 78.92 N (approximately)
Therefore, the tension in the string is approximately 78.92 N.
A force of 50 N is applied tangentially to the rim of disk 20cm in radius and mass 30kg .find the angular acceleration of the disk?
Good.......
tension = sum of centrietal force+gravity
notice these are vectors.
tensionvertical= mg down
tensionhorizonal= m w^2 * rad
= .5*(2PI/.4)^2 * .8
now, after calculating that, you add them as vectors
tension= sqrt ( (mg)^2+(horizonalabove)^2 )