A particle P of mass 2kg is attached to one end of a light rod of length 0.5m which is free to rotate in a vertical plane about its other end.The particle describes complete circles, Given that the tension at the bottom is 5times the tension at the highest point, find spped of P ar lowest point
To find the speed of particle P at the lowest point, we can use the principles of circular motion.
1. First, let's consider the forces acting on the particle at the highest point and at the lowest point of the circular motion.
At the highest point:
- The tension (T1) in the rod acts downward.
- The weight of the particle (mg) acts downward.
- The net force acting on the particle is the tension minus the weight, resulting in the centripetal force required for circular motion.
At the lowest point:
- The tension (T2) in the rod acts upward.
- The weight of the particle (mg) acts downward.
- The net force acting on the particle is the tension plus the weight, again resulting in the centripetal force required for circular motion.
2. We can write the equations for the net force acting on the particle at both the highest and lowest points:
At the highest point:
T1 - mg = (mv^2) / r
(Tension in the rod minus the weight equals the centripetal force)
At the lowest point:
T2 + mg = (mv^2) / r
(Tension in the rod plus the weight equals the centripetal force)
3. Given that the tension at the highest point (T1) is 1/5th of the tension at the lowest point (T2), we can write:
T1 = (1/5)T2
4. Now, let's substitute the value of T1 in the equation for the highest point:
(1/5)T2 - mg = (mv^2) / r
5. Substituting the value of T2 from equation 4 into equation 5:
(1/5)(mv^2) / r - mg = (mv^2) / r
6. Simplifying equation 6:
(1/5)v^2 - g = v^2
7. Rearranging equation 6 to solve for v (speed of particle P at the lowest point):
v^2 - (1/5)v^2 = g
(4/5)v^2 = g
v^2 = (5/4)g
v = √((5/4)g)
8. Plugging in the value of g (acceleration due to gravity, approximately 9.8 m/s^2):
v = √((5/4)(9.8))
v = √(12.25)
v = 3.5 m/s
Therefore, the speed of particle P at the lowest point is 3.5 m/s.
To find the speed of particle P at the lowest point, we need to analyze the forces acting on the particle at that position.
At the lowest point, the tension in the rod will be at its maximum value. Let's denote this tension as T_max.
Let's break down the forces acting on the particle at the lowest point:
1. Gravitational force (mg): This force acts vertically downward and has a magnitude equal to the product of the mass of the particle (m) and the acceleration due to gravity (g). So, the gravitational force is given by mg = 2 kg * 9.8 m/s^2 = 19.6 N.
2. Tension in the rod (T_max): This force acts radially inward, towards the center of the circular path described by the particle P. Given that the tension at the bottom is 5 times the tension at the highest point, we can write T_max = 5 * T_highest.
At the highest point, the only forces acting on the particle are the gravitational force and tension in the rod. These forces can be broken down into their vertical and horizontal components. The vertical component of the tension balances the gravitational force, and the horizontal component provides the necessary centripetal force for circular motion.
At the lowest point, the horizontal component of the tension does not contribute to the motion because it acts perpendicular to the direction of motion. Hence, the only force acting on the particle in the direction of motion is the gravitational force. Therefore, the speed of the particle at the lowest point remains the same as at the highest point.
So, to find the speed of particle P at the lowest point, we need to determine the speed at the highest point.
The tension at the highest point (T_highest) can be found using the centripetal force equation:
T_highest - mg = (mv^2) / r
where v is the speed of the particle and r is the radius of the circular path, which is equal to the length of the rod (0.5m). Rearranging the equation, we get:
T_highest = (mv^2) / r + mg
Since the tension at the bottom is 5 times T_highest, we can write:
T_max = 5 * ((mv^2) / r + mg)
To solve for v, we need to substitute this expression for T_max into the equation for centripetal force:
T_max - mg = (mv^2) / r
Making the substitution, we have:
5 * ((mv^2) / r + mg) - mg = (mv^2) / r
Simplifying the equation, we get:
5mv^2 + 5mgr - 5mg = mv^2
Rearranging, we have:
4mv^2 = 5mgr
Dividing by 4m, we get:
v^2 = (5gr) / 4
Finally, taking the square root of both sides gives us:
v = √((5gr) / 4)
Substituting the values of g (acceleration due to gravity) and r (length of the rod), we find:
v = √((5 * 9.8 m/s^2 * 0.5 m) / 4)
v = √(24.5 m^2/s^2)
v ≈ 4.95 m/s
Therefore, the speed of particle P at the lowest point is approximately 4.95 m/s.