Question

1) Suppose a pendulum bob is suspended from a pivot by a massless string of length l, measured to the center of the bob.

a) Show that if I draw pendulum back by an angle theta, the center of the bob is elevated by the distance

$\Delta h=l(1-costheta)=2l\sin ^{2}(theta/2)$

b) Using the results of part (a), show that the speed of the pendulum bob at the bottom of its travel should be

$2\sqrt{gl}sin(theta/2)$

Hint: Recall that (1-cos(theta))/2= $\sin ^{2}(theta/2))$

2) (Ramp section) Suppose the height of the ramp is h1=0.3m. and the foot of the ramp is horizontal, and is h2=1.00m above the floor. What will be the horizontal distance traveled by the following four objects before they hit the floor? Assume that R=12.7mm in each case; assume that the density of steel is $7.8gm/cm^{3}$ , and assume that the density of aluminum is $2.7gm/cm^{3}$ .

a) A solid steel sphere sliding down the ramp without friction.

b) A solid steel sphere rolling down the remp without slipping.

c) A spherical steel shell with shell thickness 1.0mm rolling down the ramp without slipping.

d) A solid aluminum sphere rolling down the ramp without slipping.

Answer 1

theta l*cos 23 theta) 93 from fig , h = L-L*cos(theta)

1) L = length of pendulum

h = distance by which bob get elevated

fron fig:

h = L- L*cos( theta)

h = L( 1- cos( theta))

[ from cos ( theta) = 1- 2*sin² ( theta/2) ]

h = L*2*sin² ( theta/2)

h = 2 L * sin² ( theta/2)

----------------------------------------------------------

b) at height h , energy = potential energy = mgh

at bottom , energy = kinetic energy = (1/2)*m*V²

m = mass of bob

V = velocity of bob at bottom

from conservation of energy:

P.E = K.E

mgh = (1/2)*m*V²

V = ( 2 g h)^1/2

= ( 2*g*2 L * sin² ( theta/2) )^1/2

= 2*( gL)^1/2 * sin(theta/2)