What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro
m the vertical? Express your answer in terms of some or all of the variables m , L, and θ, as well as the free-fall acceleration g.
1 answer:
Answer:
v = √(Lsinθ tanθ)
Explanation:
From the diagram attached,
v = the tangential speed.
r = the radius of the horizontal circle.
T = tension in the string.
θ = the angle that the string makes with the vertical
m = Bob's mass (mg = the weight)
F = centripetal force
L = the length of the string
From geometry,
r = Lsin θ
Thus, the centripetal acceleration is given as;
a = v²/r = v²/Lsin θ
Force = mass x acceleration
Thus,
Centripetal force = mv²/Lsin θ
Let's balance the vertical forces to obtain,
T cosθ - mg = 0
Thus, T cosθ = mg - - - - (eq1)
Similarly, let's balance the horizontal forces to obtain;
T sinθ = F = mv²/Lsin θ
So, T sinθ = mv²/Lsin θ - - - - (eq2)
Let's divide eq 2 by eq 1 to get;
Tsinθ/Tcosθ = (mv²/Lsin θ)/mg
tanθ = gv²/Lsinθ
Thus, v² = Lsinθ tanθ
v = √(Lsinθ tanθ)
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Answer:
b) It is impossible to tell without knowing the masses.
Explanation:
The temperature change of a substance when it receives/gives off a certain amount of heat Q is given by
where
Q is the amount of heat
m is the mass of the substance
Cs is the specific heat capacity of the substance
In this case, we have a hot piece of aluminum in contact with a cold piece of copper: the amount of heat given off by the aluminum is equal to the amount of heat absorbed by the copper, so Q is the same for the two substances. However, we see that the temperature change of the two substances depends on two other factors: the mass, m, and the specific heat, Cs. So, since we know only the specific heat of the two substances, but not their mass, we can't tell which object will experience the greater temperature change.
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.