Question

1. For a stationary ball of mass m = 0.200 kg hanging from a
massless string, draw arrows (click on the “Shapes” tab) showing
the forces acting on the ball (lengths can be arbitrary, but get
the relative lengths of each force roughly correct). For this case
of zero acceleration, use Newton’s 2^{nd} law to find the
magnitude of the tension force in the string, in units of Newtons.
Since we will be considering motion in the horizontal xy plane, we
will take the z axis to be the vertical direction. The weight w =
mg is in the downward z direction, while the tension T is in the
upward direction of the string. The vector sum of these forces
should be set equal to the mass times the acceleration of the ball,
but that is zero in this case.

2. Horizontal circular motion The ball is now set into circular
motion with a constant velocity v = 2.00 m/s and radius r = 0.300 m
in the horizontal xy plane. Draw arrows showing the forces acting
on the ball, and the direction of the centripetal acceleration.
Compute the magnitude of the acceleration, and also the period
T_{p} of the motion (the time for one complete revolution
around the circle).

3. To find the angle φ corresponding to the given velocity and
radius we need to consider the x and z components of the forces in
question 2, separately. Since there is no motion in the z
direction, the acceleration a_{z} = 0. Hence the sum of the
z components of the forces must be equal to zero. From this, solve
for an equation relating the tension force in the string to the
angle φ.

4. Take the x axis to be centered on the ball, and pointing
radially inward toward the center of the circle. Newton’s
2^{nd} law now states that the sum of the x components of
the forces in question 2 must be equal to the mass times the
centripetal acceleration. Solve this equation for the tension force
in the string, and since this must be equal to the equation you
found in slide 4, find an equation relating the angle φ to the
velocity and radius. Calculate φ for the values given on question 2
, and then from the equation for

T on question 2 find the tension in the string for the rotating
ball .

5. Vertical circular motion The ball is now set into circular motion in an xz plane perpendicular to the ground. In the drawing below, draw and label arrows showing the forces acting on the ball at the four different positions around the circle. Also show the net acceleration direction at each position (adding the vectors for the gravitational acceleration and the centripetal acceleration). The ball will slow down going up, and speed up coming down, shown by the velocity arrows in the drawing.

Answer #1

A small ball of mass 61 g is suspended from a string of length
62 cm and whirled in a circle lying in the horizontal plane. If the
string makes an angle of 25◦ with the vertical, find the
centripetal force experienced by the ball. The acceleration of
gravity is 9.8 m/s 2 . Answer in units of N

Question 1
The study of Uniform Circular Motion relates to objects
traveling with constant speed around a circle with radius, R. Since
the object has a constant speed along a circular path, we can also
say that
A)
the object has zero velocity.
B)
the object has a constant acceleration magnitude.
C)
the object has zero acceleration.
D)
the object has a constant velocity.
E)
the object has an increasing acceleration.
Question 2
Uniform Circular Motion (UCM) problems are just...

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(b) Can the strong guy exert a force to keep the string
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b) A force is necessary to change the motion of an object.
c) If an object is stationary, then no forces are acting on
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d) An object in motion must have an acceleration.
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A.) Taking the x axis to be the original direction of
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A ball with mass M=1 kg is hanging from a wire with no mass and
length l=10 cm. The wire is tilted to angle α=30 degrees. It then
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a) What is the velocity of ball V=? right before it hits the
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