an object of mass 10 kg is whirled round a horizontal circle of radius 4m by a revolving string inclined to the vertical . if the uniform speed is 5 m/s. calculate the tension in the string.
2.the angle of inclination of the string.
To calculate the tension in the string, we need to consider the forces acting on the object.
1. Centripetal Force: The tension in the string provides the centripetal force to keep the object moving in a circular path. This force is given by the formula:
Fc = (m * v^2) / r
Where:
Fc is the centripetal force
m is the mass of the object (10 kg)
v is the speed of the object (5 m/s)
r is the radius of the circle (4 m)
Plugging in the values:
Fc = (10 kg * (5 m/s)^2) / 4 m
= 125 N
Therefore, the tension in the string is 125 N.
2. Angle of Inclination: The angle of inclination of the string can be determined using trigonometry. Let's call the angle of inclination theta.
In the given problem, the string is inclined to the vertical. This means that the tension in the string can be decomposed into two components: one in the vertical direction and one in the horizontal direction.
The vertical component of tension balances the weight of the object, and the horizontal component provides the centripetal force.
Using trigonometry, we can relate the angle of inclination to the horizontal component of tension.
sin(theta) = horizontal component of tension / tension in the string
Rearranging the equation:
horizontal component of tension = tension in the string * sin(theta)
Since we know the tension in the string (125 N) and we want to find theta, we can rearrange again:
sin(theta) = horizontal component of tension / tension in the string
Plugging in the values:
sin(theta) = 125 N / 125 N
= 1
To find the angle theta, we take the inverse sine (or arcsine) of 1:
theta = arcsin(1)
= 90 degrees
Therefore, the angle of inclination of the string is 90 degrees.