Tuesday

October 21, 2014

October 21, 2014

Posted by **KimK** on Thursday, October 9, 2008 at 12:26pm.

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!

- Phys. pendulum -
**bobpursley**, Thursday, October 9, 2008 at 2:33pmYes, 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.

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