What Is the Speed of an Ice Flake in a Frictionless Hemispherical Bowl?

[mb85]

In Fig. 8-30, a 4.5 g ice flake is released from the edge of a hemispherical bowl whose radius r is 28 cm. The flake-bowl contact is frictionless. (a) What is the speed of the flake when it reaches the bottom of the bowl? (b) If we substituted a second flake with 4 times the mass, what would its speed be? (c) If, instead, we gave the flake an initial downward speed 1.4 m/s along the bowl, what would the answer be?

I know for part a.
mgh = 1/2mv^2
v = Sqaure root (2gh)
v = 2.34 m/s

part b i know is the same answer as a, because the mass does not effect the speed.

but for part c, i am not getting the right answer and it seems like a relatively basic step. can someone help me out?


Also, this problem is giving me headaches.
Figure 8-34 shows a pendulum of length L = 1.4 m. Its bob (which effectively has all the mass) has speed v0 when the cord makes an angle 0 = 48° with the vertical. (a) What is the speed of the bob when it is in its lowest position if v0 = 9.9 m/s? What is the least value that v0 can have if the pendulum is to swing down and then up (b) to a horizontal position, and (c) to a vertical position with the cord remaining straight?
http://edugen.wiley.com/edugen/courses/crs1062/art/qb/qu/c08/Fig08_34.gif

i was trying to find the y-component.
by doing cos48 = y/1.4
y = .94m

then sin 48 = x/1.4
x = 1.04m

1.4 - .94 = 0.46m
then i used v = Square root ((2)(9.8)(.46)
v = 9.016m/s
But then where do i use the initial velocity @?

im just a bit confused. thanks for ur help!

 

[Hootenanny]

For your problem, you need to factor in the intial kinetic energy to you equation.

 

[vaishakh]

mgh = 1/2m(v^2 - 1.4m/s^2) sonce it had that initial Kinetic energy before starting to slide. That it, ystem gains that much more Kinetic energy than a.

For the second question, you know the intitial Kinetic energy. Now the potential energy goes on changing and thus the Kinetic energy changes. To start the chane in potential energy for a is mg(1-cos(theta)).
Do the rest using the samee concept.

 

[mb85]

(c) If, instead, we gave the flake an initial downward speed 1.4 m/s along the bowl, what would the answer be?

mgh = 1/2m(2.34m/s^2 - 1.4m/s^2)
mgh = 7.9101

??

 

[topsquark]

mgh = 1/2m(2.34m/s^2 - 1.4m/s^2)
mgh = 7.9101

??

Think of it this way...
We have no friction so there is no nonconservative work done in the system. Thus:
$$0= \Delta PE + \Delta KE$$
Choose the bottom of the bowl as your 0 for GPE. Thus the original height of the ice flake is at h. The final point is when the flake is at zero height.
$$0=(0-mgh)+(1/2)m(v^2-v_0^2)$$
or, as vaishakh said:
$$mgh=(1/2)m(v^2-v_0^2)$$.

At this point we know $$v_0=1.4 \, m/s$$ but we don't know v anymore! All we know is that it should be greater than the 2.34 m/s you found in part a). So solve the equation for v.

-Dan