A pendulum consists of a mass of 3 kg tied to the end of a string of length 1.5 m.
The pendulum is initially held at angle of 45o with the vertical and the pendulum
is released. At the instant the mass passes through its lowest position it is 2 m
above the ground and the string breaks.
Show that the speed of the mass as it passes through its lowest position is
2.93 m/s.
(ii) Find the tension in the string just before it breaks.
(iii) Find the time it takes for the mass to hit the ground.
(iv) Find the horizontal distance traveled by the mass after the string breaks.
__________ so far, this is what i did
m = 3 kg
l = 1.5m
angle = 45o
i said, F=ma which is 3 x 9.81 = 29.43
mv^2 / r = 29.43sin45
since the height is 2m above the ground at the lowest point, r would be 1m...right?
so
3v^2 = 25
v = 2.89
however, the ans is suppose to be 2.93
for 2!
i said that Tension (T) - Weight (W) = mv2/r
found the weight by w=mg which gave 29.43
T - 29.43 = 8.79
i got 38 and the ans is suppose to be 46
i think something is wrong with the reasoning.l..HLP!
Yes, your reasoning is off.
You need to find the PE of the pendulum above the lowest point.
If R is 1.5 given, then the position of the mass vertically below the piviot point must be 1.5cos45=1.06 . So the height from the base of the swing must be 1.5-1.06=.44m
PE=mg(.44)
That has to be the KE at the bottom.
1/2 mv^2=mg(.44)
now solve for v
Now tension is mg+mv^2/r
the drop is free fall, so figure time for a 2 meter fall. THen use that time to figure the distance it went horizontally, from t and v.
To solve these problems, we can use the principles of conservation of energy and dynamics. Let's break each part down to find the correct answers:
(i) Finding the speed of the mass as it passes through its lowest position:
The potential energy at the highest point is converted into kinetic energy at the lowest point. We can equate these two energies:
Potential energy at highest point = Kinetic energy at lowest point
mgh = 0.5mv^2
Here, m is the mass (3 kg), g is the acceleration due to gravity (9.81 m/s^2), h is the vertical height (2 m), and v is the speed.
Substituting the given values:
3 * 9.81 * 2 = 0.5 * 3 * v^2
v^2 = (3 * 9.81 * 2) / 3
v^2 = 39.24
v = √39.24 ≈ 6.26 m/s (not 2.89 as you calculated)
(ii) Finding the tension in the string just before it breaks:
When the string is about to break, the centripetal force exerted by the tension is equal to the weight of the mass.
Centripetal Force = Weight
T = mg
where m is the mass (3 kg) and g is the acceleration due to gravity (9.81 m/s^2).
Substituting the given values:
T = 3 * 9.81
T ≈ 29.43 N (not 38 as you calculated)
(iii) Finding the time it takes for the mass to hit the ground:
To find the time, we can use the equation for vertical distance traveled during free fall:
h = (1/2)gt^2
where h is the height (2 m), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.
Substituting the given values:
2 = (1/2) * 9.81 * t^2
t^2 = (2 * 2) / 9.81
t^2 ≈ 0.407
t ≈ √0.407 ≈ 0.64 s
(iv) Finding the horizontal distance traveled by the mass after the string breaks:
After the string breaks, the mass will travel in a projectile motion with an initial horizontal velocity of 0 m/s.
We know that the time of flight is twice the time it takes to reach the highest point, so the total time of flight is approximately 2 * 0.64 s = 1.28 s.
The horizontal distance traveled can be calculated using the formula:
Horizontal distance = (initial horizontal velocity) * time of flight
Horizontal distance = 0 * 1.28
Horizontal distance = 0 m
Therefore, after the string breaks, the mass does not travel horizontally and falls straight to the ground.