# Inertia physics help

## Homework Statement

A sphere and a cylinder of equal mass and radius are simultaneously released from rest on the same inclined plane and roll without sliding down the incline. Then:

A. the sphere reaches the bottom first because it has the greater inertia
B. the cylinder reaches the bottom first because it picks up more rotational energy
C. the sphere reaches the bottom first because it picks up more rotational energy
D. they reach the bottom together
E. none of the above are true

I have no idea where to even start to solve this.

A hoop, a uniform disk, and a uniform sphere, all with the same mass and outer radius,
start with the same speed and roll without sliding up identical inclines. Rank the o
according to how high they go, least to greatest.

A) hoop, disk, sphere
B) disk, hoop, sphere
C) sphere, hoop, disk
D) sphere, disk, hoop'
E) hoop, sphere, disk

I know:
hoop: I=mr2
disk: I = 1/2 mr2
sphere: I = 2/5mr2

since hoop has greatest rotational kinetic energy, it will go the highest?

9. A particle is in simple harmonic motion with period T. At time t = 0 it is halfway between
the equilibrium point and an end point of its motion, traveling toward the end point. The next
time it is at the same place is:
A. t = T
B. t = T/2
C. t = T/4
D. t = T/8
E. none of the above

46. A simple pendulum consists of a small ball tied to a string and set in oscillation. As the
pendulum swings the tension force of the string is:
A. constant
B. a sinusoidal function of time
C. the square of a sinusoidal function of time
D. the reciprocal of a sinusoidal function of time
E. none of the above

my answer is B, since tension is greatest at the bottom of the swing and it oscillates so the tension when displayed on a F vs T graph it goes up and down like a sine function.

Last edited:

haruspex
Homework Helper
Gold Member
A. the sphere reaches the bottom first because it has the greater inertia
B. the cylinder reaches the bottom first because it picks up more rotational energy
C. the sphere reaches the bottom first because it picks up more rotational energy
D. they reach the bottom together
E. none of the above are true

I have no idea where to even start to solve this.
Start by quoting the moment of inertia of each. Will a higher moment make the object go faster or slower?
A hoop, a uniform disk, and a uniform sphere, all with the same mass and outer radius,
start with the same speed and roll without sliding up identical inclines. Rank the o
according to how high they go, least to greatest.

A) hoop, disk, sphere
B) disk, hoop, sphere
C) sphere, hoop, disk
D) sphere, disk, hoop'
E) hoop, sphere, disk

I know:
hoop: I=mr2
disk: I = 1/2 mr2
sphere: I = 2/5mr2

since hoop has greatest rotational kinetic energy, it will go the highest?
9. A particle is in simple harmonic motion with period T. At time t = 0 it is halfway between
the equilibrium point and an end point of its motion, traveling toward the end point. The next
time it is at the same place is:
A. t = T
B. t = T/2
C. t = T/4
D. t = T/8
E. none of the above

Write down a generic equation for SHM. At what t values is it at half amplitude?
46. A simple pendulum consists of a small ball tied to a string and set in oscillation. As the
pendulum swings the tension force of the string is:
A. constant
B. a sinusoidal function of time
C. the square of a sinusoidal function of time
D. the reciprocal of a sinusoidal function of time
E. none of the above

my answer is B, since tension is greatest at the bottom of the swing and it oscillates so the tension when displayed on a F vs T graph it goes up and down like a sine function.
Yes, but functions other than sine can look like that.
Can you derive an equation for the tension?

Start by quoting the moment of inertia of each. Will a higher moment make the object go faster or slower?

cylinder: I = 1/2 mr2
sphere: I = 2/5mr2
is rotational inertia same as moment of inertia?

okay i did some calculations and this is what i got:

sphere: v2=(10/7)*g*h
cylinder: v2=(4/3)*g*h

so the sphere has the greatest velocity and should reach the bottom first?

and now with that, comes another question:

Two uniform cylinders have different masses and different rotational inertias. They simultaneously
start from rest at the top of an inclined plane and roll without sliding down the plane.
The cylinder that gets to the bottom first is:
A. the one with the larger mass
B. the one with the smaller mass
C. the one with the larger rotational inertia
D. the one with the smaller rotational inertia
E. neither (they arrive together)

the answer is: the one with the smaller rotational inertia reaches the bottom first?
but at the same time... we dont have mass so we cant really determine it?

Write down a generic equation for SHM. At what t values is it at half amplitude?

x(t) = Acos(ωt)
A/2 = Acos(ωt)
1/2 = cos(ωt)
ωt = ∏/3 and 5∏/3?

and
1)
2∏f*t = ∏/3
2f*t = 1/3
f*t = 1/6
(1/T)*t = 1/6
t = T/6?

2)
(2∏/T)*t = 5∏/3
t = 5T/6?

Yes, but functions other than sine can look like that.
Can you derive an equation for the tension?

i have no idea how to do that ahh. i have my final exam in 2 hours

i only know that:
mv2/r = T - Fgcosθ?

Last edited:
haruspex
Homework Helper
Gold Member
cylinder: I = 1/2 mr2
sphere: I = 2/5mr2
is rotational inertia same as moment of inertia?
Yes.
Yes.
x(t) = Acos(ωt +3∏/2)? is this the correct equation?
Anything of the form Acos(ωt), Asin(ωt), Acos(ωt+β), ... The phase won't matter because we're only interested in time differences, not an absolute time.
A/2 = ?
Take e.g. x= Asin(ωt). For what values of t will x = A/2?
mv2/r = T - Fgcosθ?
Thinking about this again, it seems to me that options B and C should be interpreted literally as T ~ sin(ωt+β) and T ~ sin2(ωt+β). That means you can rule those out on the basis that it's never zero. And you can rule out D because it's never infinite.
Using your equation (but I think you meant mgcosθ), you also know how v and θ vary as a function of time, right? So that gives you an equation for T as a function of time.
Now, as we know, a pendulum is only approximately SHM, so I'm going to assume that the same degree of approximation is intended in answering this question. And since you're short of time, I'll break the rules and just go ahead with an exposition.
θ = Asin(ωt) (say) where ω2=g/L
v = Aωcos(ωt)L
cos(θ) ≈ 1 - θ2/2
T = mA2ω2L2cos2(ωt)/L + mg(1-A2sin2(ωt)/2)
= mg{1+A2cos2(ωt)-A2sin2(ωt)/2}
which can be written in the form C + D*sin2(ωt).
So clearly it is not a constant.
(That said, I'm surprised the answer is quite so messy, so I may have made a mistake. It would only take an error in a sign and one in a factor of two to end up with T = constant, but I feel intuitively that wouldn't be right.)
Now, if the options B and C really meant to allow for T ~ constant + sin(ωt+β) or T ~ constant + sin2(ωt+β), then you might say the answer is C. However, you could further rewrite the expression as C' + D'*sin(2ωt+β). So if you take that interpretation then options B and C are the same.

Yes.

Yes.
cylinder: I = 1/2 mr2
sphere: I = 2/5mr2
is rotational inertia same as moment of inertia?

okay i did some calculations and this is what i got:

sphere: v2=(10/7)*g*h
cylinder: v2=(4/3)*g*h

so the sphere has the greatest velocity and should reach the bottom first?

and now with that, comes another question:

Two uniform cylinders have different masses and different rotational inertias. They simultaneously
start from rest at the top of an inclined plane and roll without sliding down the plane.
The cylinder that gets to the bottom first is:
A. the one with the larger mass
B. the one with the smaller mass
C. the one with the larger rotational inertia
D. the one with the smaller rotational inertia
E. neither (they arrive together)

the answer is: the one with the smaller rotational inertia reaches the bottom first?
but at the same time... we dont have mass so we cant really determine it?

Anything of the form Acos(ωt), Asin(ωt), Acos(ωt+β), ... The phase won't matter because we're only interested in time differences, not an absolute time.

Take e.g. x= Asin(ωt). For what values of t will x = A/2?
x(t) = Acos(ωt)
A/2 = Acos(ωt)
1/2 = cos(ωt)
ωt = ∏/3 and 5∏/3?

and
1)
2∏f*t = ∏/3
2f*t = 1/3
f*t = 1/6
(1/T)*t = 1/6
t = T/6?

2)
(2∏/T)*t = 5∏/3
t = 5T/6?

difference in time = 4T/6 = 2T/3?

so answer is none of the above?

Thinking about this again, it seems to me that options B and C should be interpreted literally as T ~ sin(ωt+β) and T ~ sin2(ωt+β). That means you can rule those out on the basis that it's never zero. And you can rule out D because it's never infinite.
Using your equation (but I think you meant mgcosθ), you also know how v and θ vary as a function of time, right? So that gives you an equation for T as a function of time.
Now, as we know, a pendulum is only approximately SHM, so I'm going to assume that the same degree of approximation is intended in answering this question. And since you're short of time, I'll break the rules and just go ahead with an exposition.
θ = Asin(ωt) (say) where ω2=g/L
v = Aωcos(ωt)L
cos(θ) ≈ 1 - θ2/2
T = mA2ω2L2cos2(ωt)/L + mg(1-A2sin2(ωt)/2)
= mg{1+A2cos2(ωt)-A2sin2(ωt)/2}
which can be written in the form C + D*sin2(ωt).
So clearly it is not a constant.
(That said, I'm surprised the answer is quite so messy, so I may have made a mistake. It would only take an error in a sign and one in a factor of two to end up with T = constant, but I feel intuitively that wouldn't be right.)
Now, if the options B and C really meant to allow for T ~ constant + sin(ωt+β) or T ~ constant + sin2(ωt+β), then you might say the answer is C. However, you could further rewrite the expression as C' + D'*sin(2ωt+β). So if you take that interpretation then options B and C are the same.
so the answer i nthe book is wrong? it says its E.

haruspex
Homework Helper
Gold Member
so the sphere has the greatest velocity and should reach the bottom first?
Yes. As the PE is turned into KE, the object with the least MI gets the greatest fraction of its KE in the form of translational KE.
Two uniform cylinders have different masses and different rotational inertias. They simultaneously start from rest at the top of an inclined plane and roll without sliding down the plane. The cylinder that gets to the bottom first is:
A. the one with the larger mass
B. the one with the smaller mass
C. the one with the larger rotational inertia
D. the one with the smaller rotational inertia
E. neither (they arrive together)

the answer is: the one with the smaller rotational inertia reaches the bottom first?
but at the same time... we dont have mass so we cant really determine it?
Don't just guess. Write out some equations and deduce the answer.
x(t) = Acos(ωt)
A/2 = Acos(ωt)
1/2 = cos(ωt)
ωt = ∏/3 and 5∏/3?
t = T/6, 5T/6?
difference in time = 4T/6 = 2T/3?
so answer is none of the above?
Yes, that's enough to choose "none". But to work out the required phase difference correctly you'd need to consider that -∏/3 etc. are also solutions. In between consecutive solutions, the object will either reach maximum deviation or zero deviation. You are asked for an interval during which it reaches max deviation. That will be either 2T/3 or T/3 (in fact it's the second), so the multiple choice answer is still "none".
so the answer in the book is wrong? it says its E.
No, read what I wrote again:
"it seems to me that options B and C should be interpreted literally as T ~ sin(ωt+β) and T ~ sin2(ωt+β). That means you can rule those out on the basis that it's never zero. And you can rule out D because it's never infinite."
I then went on to show that (unless I'd made a mistake in the algebra) it was not constant either. That leaves E.

ehild
Homework Helper
Two uniform cylinders have different masses and different rotational inertias. They simultaneously
start from rest at the top of an inclined plane and roll without sliding down the plane.
The cylinder that gets to the bottom first is:
A. the one with the larger mass
B. the one with the smaller mass
C. the one with the larger rotational inertia
D. the one with the smaller rotational inertia
E. neither (they arrive together)

the answer is: the one with the smaller rotational inertia reaches the bottom first?
but at the same time... we dont have mass so we cant really determine it?

If uniformity means that the density of the cylinder is constant, the moment of inertia is 0.5mr2. The kinetic energy of the cylinder is KE=1/2(mv2+0.5mr2ω2) and v=rω in case of rolling, so KE=(3/4)mv2
At height h, KE+mgh=mgh0,
v2=(4/3)g(h0-h).
The cylinders have the same speed at the same height, independently of their mass or moment of inertia.

ehild