WY5V8181
HR

(a)

Graph B is correct because the magnitude of acceleration is constant in
simple circular motion, so the magnitude of net force is proportional
and thus constant (F = ma).

(b)

The magnitude of tension on the string is greater than the centripetal
force on the sphere. Tension has two components, the opposite force to
gravity because the sphere is in equilibrium vertically, and the
centripetal force. The resultant of two nonzero forces is larger than
either component, so tension > centripetal force.

(c)

The inward acceleration needn't be as large for a very small tangential
speed to keep the sphere in constant circular motion. This means that
the centripetal force can be smaller. If the centripetal force can be
smaller but the vertical force of tension counteracting gravity is the
same, then the angle of the string and the pole is smaller.

(d)

This equation is consistent with part (c) because, for small values of
theta (string nearly vertical), tan(theta)*sin(theta) ~ 0, so v^2 ~ 0,
so v ~ 0 (small).

(e)

v^2 = gL*tan(theta)*sin(theta)
=> g = v^2 / (tan(theta) * sin(theta)).

If v^2 were graphed on the y-axis of a graph and
(tan(theta) * sin(theta)) on the x-axis, the slope of the line of best
fit would give an experimental value of g.

(f)

As theta decreases, the distance of the sphere to the rod decreases.
The masses of all objects remains const, I = mr^2 for a point-like
object, and the distances from the pivot (and thus inertia) remain the
same for the rod and the platform. Therefore, as r -> 0, I -> 0.