A child of mass m swings in a swing supported by two chains, each of length R.
The mass of the swing’s seat is negligible. If the tension in each chain at the
lowest point is T, find:
4.1 the child’s speed at the lowest point;
4.2 the force exerted by the seat on the child at the lowest point.
To find the child's speed at the lowest point and the force exerted by the seat on the child, we can use the concepts of circular motion and centripetal force.
4.1 The child's speed at the lowest point:
At the lowest point of the swing, the tension in the chains provides the centripetal force required to keep the child in circular motion. The tension in the chains acts towards the center of the circular path.
The centripetal force (Fc) is given by the formula:
Fc = (m * v^2) / R
Where m is the mass of the child, v is the child's speed, and R is the length of the chains.
At the lowest point, the tension in each chain is T, which provides the centripetal force. So we have:
T = (m * v^2) / R
To find the child's speed (v), we can rearrange the equation:
v^2 = (T * R) / m
v = sqrt((T * R) / m)
Therefore, the child's speed at the lowest point is v = sqrt((T * R) / m).
4.2 The force exerted by the seat on the child:
The seat exerts a normal force on the child, which is equal in magnitude but opposite in direction to the force exerted by the child on the seat. At the lowest point, the child experiences both the tension force from the chains and the gravitational force acting downwards.
The net force acting on the child is the vector sum of these two forces. The net force is given by:
Fnet = T + mg
Where T is the tension in each chain, m is the mass of the child, and g is the acceleration due to gravity.
The force exerted by the seat on the child is equal in magnitude but opposite in direction to the net force. Therefore, the force exerted by the seat can be calculated as:
Force exerted by the seat = -Fnet = -(T + mg)
Therefore, the force exerted by the seat on the child at the lowest point is -(T + mg).
Note: The negative sign indicates that the force exerted by the seat is in the opposite direction to the net force.
To find the child's speed at the lowest point, we can analyze the forces acting on the child.
At the lowest point of the swing, the child's weight acts downwards and the tension in the chains acts upwards. The net force acting on the child is the centripetal force required to keep the child in circular motion.
4.1 To find the child's speed at the lowest point, we can equate the net force to the centripetal force.
Net force = Tension - Weight
Centripetal force = (mass × velocity^2) / radius
Since the child is at the lowest point, the tension in each chain is T and the weight is mg (mass × acceleration due to gravity).
Setting the net force equal to the centripetal force, we get:
T - mg = (m × v^2) / R
Simplifying the equation, we have:
v^2 = (T - mg) × R / m
Taking the square root of both sides, we get:
v = √((T - mg) × R / m)
So, the child's speed at the lowest point is √((T - mg) × R / m).
4.2 To find the force exerted by the seat on the child at the lowest point, we can analyze the forces in the vertical direction.
At the lowest point, there are two forces acting in the vertical direction: the tension in each chain (T) and the weight of the child (mg).
The force exerted by the seat on the child is equal to the net force acting in the vertical direction.
Net force = T - mg
So, the force exerted by the seat on the child at the lowest point is T - mg.
Lets swing a ball on the end of ONE string and divide call it 2 T
You were not told what angle or height so I will say dropped from height h.
m g h = (1/2) m v^2
so at the bottom v = sqrt (2gh)
then the centripetal acceleration = v^2/R
and 2 T = m g + m v^2/R
2 T= m (g + 2 g h/R) = mg (1 +2 h/R) = force up on child if sat has no mass :)
Now if you were given maximum angle theta instead of height h then
h = R - R cos theta = R(1-cos theta)
if theta is small cos theta = 1 -theta^2/2 +.....
so
h = (R /2)theta^2 note theta in radians