a stone of mass 1 kg is tied to the end of a string 1m long . it is whirled in a vertical circle . if the velocity of the stone at the top be 4m/s , what is the tension in the string ?
T + mg = mv²/R
=> T = m (v²/R -g)
To determine the tension in the string when a stone of mass 1 kg is whirled in a vertical circle with a velocity of 4 m/s at the top, we can use the principles of circular motion and the laws of motion.
1. Draw a diagram: Draw a diagram of the system, representing the stone attached to the string, moving in a circular path.
_
|\
| \
| \
r | \ Tension (T)
| \
| \
|______\
Stone
2. Identify the forces: There are two forces acting on the stone at the top of the circular motion - the tension force (T) pulling the stone towards the center, and the gravitational force (mg) acting vertically downwards.
3. Apply Newton's second law: At the top of the circular motion, the net force acting on the stone is given by F_net = T - mg (where m is the mass of the stone, g is the acceleration due to gravity).
4. Apply Newton's second law in the vertical direction: Since the stone is moving in a vertical circle, use the vertical component of velocity (v_y = 4 m/s) to find the gravitational force acting on the stone. At the top of the motion, the gravitational force is given by F_gravity = mg = m * g.
5. Equate the net force to the gravitational force: Set up the equation F_net = T - mg = 0, since the stone is not accelerating vertically at the top. Therefore, T - mg = 0.
6. Solve for Tension (T): Rearranging the equation, we have T = mg. Plugging in the values, we get T = (1 kg) * (9.8 m/s²), which gives T = 9.8 N.
Therefore, the tension in the string when the stone is whirled in a vertical circle with a velocity of 4 m/s at the top is 9.8 Newtons.