Solve the equations
1/2mv2 + 1/2Iω2 = mgh and v = rω
for the speed v using substitution, given that
I = mr2
and
h = 4.20 m.
(Note that mass m and radius r will both cancel, so their numerical values aren't required.)
m/s
To solve the equations 1/2mv^2 + 1/2Iω^2 = mgh and v = rω for the speed v using substitution, we can substitute the given values of I and h.
First, we substitute the value of I: I = mr^2.
1/2mv^2 + 1/2(mr^2)ω^2 = mgh
Next, we substitute the value of h: h = 4.20 m.
1/2mv^2 + 1/2(mr^2)ω^2 = mg(4.20)
Since the mass m and radius r will cancel out, we don't need their numerical values.
Now, we can simplify the equation:
1/2v^2 + 1/2(r^2)ω^2 = g(4.20)
Using the equation v = rω, we can substitute v with rω:
1/2(rω)^2 + 1/2(r^2)ω^2 = g(4.20)
1/2(r^2ω^2) + 1/2(r^2)ω^2 = g(4.20)
Now, let's combine like terms:
1/2(r^2ω^2 + r^2ω^2) = g(4.20)
1/2(2r^2ω^2) = g(4.20)
r^2ω^2 = 2g(4.20)
Finally, we solve for v:
v = rω
v = sqrt(2g(4.20))
Please note that the numerical values for mass m and radius r are not required to solve for the speed v.
To solve the equations 1/2mv^2 + 1/2Iω^2 = mgh and v = rω for the speed v using substitution, we can follow these steps:
Step 1: Substitute the value of I in terms of m and r:
I = mr^2
Step 2: Substitute the value of h:
h = 4.20 m
Step 3: Rewrite the equations with the substitutions:
1/2mv^2 + 1/2(mr^2)ω^2 = mgh
v = rω
Step 4: Rearrange the first equation to solve for v^2:
1/2mv^2 + 1/2(mr^2)ω^2 = mgh
mv^2 + mr^2ω^2 = 2mgh
v^2 + r^2ω^2 = 2gh
Step 5: Substitute the value of v in terms of ω:
v = rω
Step 6: Substitute this value of v in the equation from step 4:
(rω)^2 + r^2ω^2 = 2gh
Step 7: Simplify the equation:
r^2ω^2 + r^2ω^2 = 2gh
2r^2ω^2 = 2gh
Step 8: Cancel out common terms:
r^2ω^2 = gh
Step 9: Solve for ω^2:
ω^2 = gh/r^2
Step 10: Take the square root of both sides to solve for ω:
ω = √(gh/r^2)
Step 11: Substitute the value of ω into the equation v = rω:
v = rω
v = r√(gh/r^2)
Step 12: Simplify the expression:
v = √(grh)
Therefore, the speed v is equal to the square root of (grh), where g represents the acceleration due to gravity and r and h represent the radius and height respectively.
This is Algebra. Surely you are not having difficulty with this in AP Physics.
1/2 mv^2+1/2 mr^2*V^2/r^2
1/2 m(v^2+v^2)=mv^2
v= sqrt(gh)
Now note something about this. If the object had just fell(no rolling)
v would have been sqrt(2gh)
What heppened, is that energy split equally into rolling, and translational motion.