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+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.