Flag Mechanics> mechanics -9...
question mark

A child of mass m sits in a swing of negligible mass suspended by a rope of length l. Assume that the dimensions of the child are negligible compared with l.
His father pulls the child back until the rope makes an angle of one radian with vertical, then pushes with a force F=mg along the arc of the circle releasing at the vertical :
a) How high up will the swing go?
b) How long did the father push?

sohan singh singh , 16 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

To tackle this problem, we need to analyze the motion of the swing and the forces acting on the child. We can break this down into two parts: determining how high the swing will go after being released and calculating the duration of the push. Let's dive into each part step by step.

Determining the Maximum Height of the Swing

When the father pushes the child, he imparts kinetic energy to the child. As the swing moves upward, this kinetic energy is converted into potential energy. The key to finding out how high the swing goes lies in the conservation of energy principle.

Energy Conservation

The total mechanical energy in the system (kinetic energy + potential energy) remains constant if we ignore air resistance and friction. Initially, when the swing is at the lowest point (the vertical position), all the energy is kinetic. At the highest point of the swing, all the energy will be potential.

  • Kinetic Energy (KE): At the lowest point, KE = (1/2)mv², where v is the velocity of the child.
  • Potential Energy (PE): At the highest point, PE = mgh, where h is the height gained and g is the acceleration due to gravity.

At the moment the swing is released from the vertical position, the potential energy at the angle of one radian can be calculated using the height gained from the vertical position:

Calculating Height

When the swing is pulled back to an angle θ (in this case, 1 radian), the height (h) can be found using the formula:

h = l - l cos(θ)

Substituting θ = 1 radian:

h = l - l cos(1) = l(1 - cos(1))

Now, we can find the maximum height the swing will reach after being pushed. The potential energy at the highest point will equal the kinetic energy imparted by the push:

mgh = (1/2)mv²

Since we know h from the previous calculation, we can rearrange this to find the maximum height:

Final Height Calculation

Setting the energies equal gives us:

v² = 2gh

Substituting h = l(1 - cos(1)) into the equation allows us to find the height the swing reaches after the push.

Duration of the Push

To find out how long the father pushed the child, we need to consider the force applied and the resulting acceleration. The force F = mg is applied along the arc of the swing.

Acceleration and Motion

When the father pushes the child, he applies a force that results in an acceleration. The net force acting on the child can be expressed as:

F_net = F - mg = ma

Since F = mg, the net force becomes:

0 = ma

This means that the child experiences no net acceleration while the force is applied. However, the child will accelerate once released. To find the time of the push, we can use the relationship between distance, velocity, and acceleration.

Calculating Time

Assuming the child starts from rest and accelerates uniformly, we can use the equation:

d = (1/2)at²

Where d is the distance along the arc of the swing. The distance can be approximated by the arc length for small angles:

d = lθ

Substituting θ = 1 radian gives us:

d = l

Now, we can find the time t by rearranging the equation:

t = √(2d/a)

Since we have the distance and can calculate the acceleration from the force applied, we can determine the duration of the push.

Summary

In summary, the maximum height the swing will reach after being pushed can be calculated using energy conservation principles, while the duration of the push can be derived from the relationship between distance, acceleration, and time. By applying these concepts, we can effectively analyze the motion of the swing and the forces at play.

ApprovedApproved
Last Activity: 10 Months ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments