A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu
1 answer:
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
Then we can calculate the angular momentum of the stone about the center of the circle, given by
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.
Substituting the data of the problem, we find
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1) The kinetic energy of an object is given by:
where m is the object's mass and v its speed.
By using this equation, we find the initial kinetic energy of the skateboarder:
and the final kinetic energy as well:
So, her change in kinetic energy is
2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
Therefore, the work done by the skateboarder is
where m is the object's mass and v its speed.
By using this equation, we find the initial kinetic energy of the skateboarder:
and the final kinetic energy as well:
So, her change in kinetic energy is
2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
Therefore, the work done by the skateboarder is
A. 6.67 cm
The focal length of the lens can be found by using the lens equation:
where we have
f = focal length
s = 12 cm is the distance of the object from the lens
s' = 15 cm is the distance of the image from the lens
Solving the equation for f, we find
B. Converging
According to sign convention for lenses, we have:
- Converging (convex) lenses have focal length with positive sign
- Diverging (concave) lenses have focal length with negative sign
In this case, the focal length of the lens is positive, so the lens is a converging lens.
C. -1.25
The magnification of the lens is given by
where
s' = 15 cm is the distance of the image from the lens
s = 12 cm is the distance of the object from the lens
Substituting into the equation, we find
D. Real and inverted
The magnification equation can be also rewritten as
where
y' is the size of the image
y is the size of the object
Re-arranging it, we have
Since in this case M is negative, it means that y' has opposite sign compared to y: this means that the image is inverted.
Also, the sign of s' tells us if the image is real of virtual. In fact:
- s' is positive: image is real
- s' is negative: image is virtual
In this case, s' is positive, so the image is real.
E. Virtual
In this case, the magnification is 5/9, so we have
which can be rewritten as
which means that s' has opposite sign than s: therefore, the image is virtual.
F. 12.0 cm
From the magnification equation, we can write
and then we can substitute it into the lens equation:
and we can solve for s:
G. -6.67 cm
Now the image distance can be directly found by using again the magnification equation:
And the sign of s' (negative) also tells us that the image is virtual.
H. -24.0 cm
In this case, the image is twice as tall as the object, so the magnification is
M = 2
and the distance of the image from the lens is
s' = -24 cm
The problem is asking us for the image distance: however, this is already given by the problem,
s' = -24 cm
so, this is the answer. And the fact that its sign is negative tells us that the image is virtual.