1. A uniform disk is rolled down a hill of height 11m. What is the speed of the disk when it reaches the bottom of the hill?
2. As soon as the disk reaches the bottom of the hill, it hits a horizontal surface of frictional coefficient 0.13. What is the acceleration of the disk?
3. A ball of mass 2kg is attached to a string of length 0.75m and rotated vertically with an angular velocity of 7rad/s. What is the ratio of the string tension at the top of the loop to the string tension at the bottom?
Ignore significant figures.

1) g * (11 m) = 0.5v^2 14.6878 m/s
2) g * 0.13 = 1.275 m/s^2 1.275 m/s^2 against the motion of the disk
3)F + (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 17.13 N
F - (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 56.37 N 0.039

1. A uniform disk is rolled down a hill of height 11m. What is the speed of the disk when it reaches the bottom of the hill?
2. As soon as the disk reaches the bottom of the hill, it hits a horizontal surface of frictional coefficient 0.13. What is the acceleration of the disk?
3. A ball of mass 2kg is attached to a string of length 0.75m and rotated vertically with an angular velocity of 7rad/s. What is the ratio of the string tension at the top of the loop to the string tension at the bottom?
Ignore significant figures.

1) g * (11 m) = 0.5v^2 14.6878 m/s
2) g * 0.13 = 1.275 m/s^2 1.275 m/s^2 against the motion of the disk
3)F + (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 17.13 N
F - (2 kg) * g = (7 rad/s * (0.75 m/rad))^2 / (0.75 m)
F = 56.37 N 0.039

The centrifugal force must be [math]\frac{mv^2}{r}[/math]. You wrote [math]\frac{v^2}{r}[/math] instead. ;)
With the fix you get [math]T_{top} = 53.9N[/math] and [math]T_{bottom} = 93.1N[/math].
Ratio ends up being 0.579.

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For the first two questions, you must take into account that the object is a rolling disk, rather than a sliding block. The disk has rotational kinetic energy as well as translational kinetic energy. Thus, the gravitational energy is equal to the sum of the rotational and translational kinetic energies.
[math]mgh=\frac{1}{2}Iw^2+\frac{1}{2}mv^2[/math]
I for a rolling solid uniform disk is given by [math]I=\frac{1}{2}mr^2[/math], where r is the radius of the disk. Thus, the mass of the disk cancels out and we are left with:
[math]gh=\frac{1}{4}r^2w^2+\frac{1}{2}v^2[/math]
By [math]v=rw[/math], [math]r^2w^2[/math] can be simplified to [math]v^2[/math], and combination of fractions gives:
[math]gh=\frac{3}{4}v^2[/math]
Algebraic manipulation gives an expression for v: [math]\sqrt{\frac{4}{3}gh}=v[/math]
Plugging in values for g and h yields [math]v=11.99\frac{m}{s}[/math]
As for the other question, if I am not mistaken, you must treat the frictional force as a torque acting tangential to the radius of the disk. [math]T=FL[/math]sin theta (idk how to use this math thing). theta is 90 and sin(90) is 1, so the torque is simply [math]FL[/math]. Using the formula for frictional force and [math]T=Iq[/math] (q is alpha or angular acceleration, again, idk how to use this math thing) gives the following equation (using u as mu for frictional coefficient):
[math]umgr=\frac{1}{2}mr^2q[/math]
m and one r again cancel out. Using [math]a=rq[/math], the other r is gone as well, leaving:
[math]ug=\frac{1}{2}a[/math] and thus [math]2ug=a[/math].
Plugging in values for u and g yields [math]a=2.548\frac{m}{s^2}[/math]
I do not think that rotational motion is allowed in the rules but I like to have a little fun. Sorry if this causes too many headaches. Please inform me if you find anything wrong with my solution.

i wish i was good
Events 2019: Expd, Water, Herp
Rip states 2019

For the first two questions, you must take into account that the object is a rolling disk, rather than a sliding block. The disk has rotational kinetic energy as well as translational kinetic energy. Thus, the gravitational energy is equal to the sum of the rotational and translational kinetic energies.
[math]mgh=\frac{1}{2}Iw^2+\frac{1}{2}mv^2[/math]
I for a rolling solid uniform disk is given by [math]I=\frac{1}{2}mr^2[/math], where r is the radius of the disk. Thus, the mass of the disk cancels out and we are left with:
[math]gh=\frac{1}{4}r^2w^2+\frac{1}{2}v^2[/math]
By [math]v=rw[/math], [math]r^2w^2[/math] can be simplified to [math]v^2[/math], and combination of fractions gives:
[math]gh=\frac{3}{4}v^2[/math]
Algebraic manipulation gives an expression for v: [math]\sqrt{\frac{4}{3}gh}=v[/math]
Plugging in values for g and h yields [math]v=11.99\frac{m}{s}[/math]
As for the other question, if I am not mistaken, you must treat the frictional force as a torque acting tangential to the radius of the disk. [math]T=FL[/math]sin theta (idk how to use this math thing). theta is 90 and sin(90) is 1, so the torque is simply [math]FL[/math]. Using the formula for frictional force and [math]T=Iq[/math] (q is alpha or angular acceleration, again, idk how to use this math thing) gives the following equation (using u as mu for frictional coefficient):
[math]umgr=\frac{1}{2}mr^2q[/math]
m and one r again cancel out. Using [math]a=rq[/math], the other r is gone as well, leaving:
[math]ug=\frac{1}{2}a[/math] and thus [math]2ug=a[/math].
Plugging in values for u and g yields [math]a=2.548\frac{m}{s^2}[/math]
I do not think that rotational motion is allowed in the rules but I like to have a little fun. Sorry if this causes too many headaches. Please inform me if you find anything wrong with my solution.

Nah, I think rotational motion is fair game. I just haven't gotten to it yet >.>

EDIT: For #1, in order for the ball (EDIT2: to roll), don't you have to apply a torque to it first though?

Last edited by UTF-8 U+6211 U+662F on April 11th, 2018, 11:49 am, edited 1 time in total.

For the first two questions, you must take into account that the object is a rolling disk, rather than a sliding block. The disk has rotational kinetic energy as well as translational kinetic energy. Thus, the gravitational energy is equal to the sum of the rotational and translational kinetic energies.
[math]mgh=\frac{1}{2}Iw^2+\frac{1}{2}mv^2[/math]
I for a rolling solid uniform disk is given by [math]I=\frac{1}{2}mr^2[/math], where r is the radius of the disk. Thus, the mass of the disk cancels out and we are left with:
[math]gh=\frac{1}{4}r^2w^2+\frac{1}{2}v^2[/math]
By [math]v=rw[/math], [math]r^2w^2[/math] can be simplified to [math]v^2[/math], and combination of fractions gives:
[math]gh=\frac{3}{4}v^2[/math]
Algebraic manipulation gives an expression for v: [math]\sqrt{\frac{4}{3}gh}=v[/math]
Plugging in values for g and h yields [math]v=11.99\frac{m}{s}[/math]
As for the other question, if I am not mistaken, you must treat the frictional force as a torque acting tangential to the radius of the disk. [math]T=FL[/math]sin theta (idk how to use this math thing). theta is 90 and sin(90) is 1, so the torque is simply [math]FL[/math]. Using the formula for frictional force and [math]T=Iq[/math] (q is alpha or angular acceleration, again, idk how to use this math thing) gives the following equation (using u as mu for frictional coefficient):
[math]umgr=\frac{1}{2}mr^2q[/math]
m and one r again cancel out. Using [math]a=rq[/math], the other r is gone as well, leaving:
[math]ug=\frac{1}{2}a[/math] and thus [math]2ug=a[/math].
Plugging in values for u and g yields [math]a=2.548\frac{m}{s^2}[/math]
I do not think that rotational motion is allowed in the rules but I like to have a little fun. Sorry if this causes too many headaches. Please inform me if you find anything wrong with my solution.

Nah, I think rotational motion is fair game. I just haven't gotten to it yet >.>

EDIT: For #1, in order for the ball, don't you have to apply a torque to it first though?

You can assume that the gravity does all of the work, like it starts on an incline.

i wish i was good
Events 2019: Expd, Water, Herp
Rip states 2019

For the first two questions, you must take into account that the object is a rolling disk, rather than a sliding block. The disk has rotational kinetic energy as well as translational kinetic energy. Thus, the gravitational energy is equal to the sum of the rotational and translational kinetic energies.
[math]mgh=\frac{1}{2}Iw^2+\frac{1}{2}mv^2[/math]
I for a rolling solid uniform disk is given by [math]I=\frac{1}{2}mr^2[/math], where r is the radius of the disk. Thus, the mass of the disk cancels out and we are left with:
[math]gh=\frac{1}{4}r^2w^2+\frac{1}{2}v^2[/math]
By [math]v=rw[/math], [math]r^2w^2[/math] can be simplified to [math]v^2[/math], and combination of fractions gives:
[math]gh=\frac{3}{4}v^2[/math]
Algebraic manipulation gives an expression for v: [math]\sqrt{\frac{4}{3}gh}=v[/math]
Plugging in values for g and h yields [math]v=11.99\frac{m}{s}[/math]
As for the other question, if I am not mistaken, you must treat the frictional force as a torque acting tangential to the radius of the disk. [math]T=FL[/math]sin theta (idk how to use this math thing). theta is 90 and sin(90) is 1, so the torque is simply [math]FL[/math]. Using the formula for frictional force and [math]T=Iq[/math] (q is alpha or angular acceleration, again, idk how to use this math thing) gives the following equation (using u as mu for frictional coefficient):
[math]umgr=\frac{1}{2}mr^2q[/math]
m and one r again cancel out. Using [math]a=rq[/math], the other r is gone as well, leaving:
[math]ug=\frac{1}{2}a[/math] and thus [math]2ug=a[/math].
Plugging in values for u and g yields [math]a=2.548\frac{m}{s^2}[/math]
I do not think that rotational motion is allowed in the rules but I like to have a little fun. Sorry if this causes too many headaches. Please inform me if you find anything wrong with my solution.

Nah, I think rotational motion is fair game. I just haven't gotten to it yet >.>

EDIT: For #1, in order for the ball, don't you have to apply a torque to it first though?

You can assume that the gravity does all of the work, like it starts on an incline.

Gravity and the normal force only apply translational motion though (as far as I know), so the only force that would apply torque is friction, which wouldn't be specified.

Nah, I think rotational motion is fair game. I just haven't gotten to it yet >.>

EDIT: For #1, in order for the ball, don't you have to apply a torque to it first though?

You can assume that the gravity does all of the work, like it starts on an incline.

Gravity and the normal force only apply translational motion though (as far as I know), so the only force that would apply torque is friction, which wouldn't be specified.

(Also I asked my q in case you didn't see it)

Momentarily consider the ball to be a rod (same mechanics) and consider the pivot point to be the contact point between the rod and the ground. Friction can be neglected entirely, otherwise I do not believe that the above method of solution works because it neglects the work done by friction (sorry, should have specified in my original post). Because the normal force on the rod acts through the pivot point, the lever arm is zero and this force can be neglected. Thus, the only force acting upon the rod is the weight of the rod, but this weight acts as a torque because it is not acting through the pivot point, which causes the rod to tip over down the incline and fall. Applying this same principle to a ball, this torque would cause the ball to begin rotating.

the force that the carriage applies on the horse is equal to the force that the horse applies on the carriage, not the total force that the horse applies. The horse applies enough force on itself and the carriage to move them both but that force is not equal to the force that the horse applies on the carriage and the force that the carriage applies on the horse.

i wish i was good
Events 2019: Expd, Water, Herp
Rip states 2019

You can assume that the gravity does all of the work, like it starts on an incline.

Gravity and the normal force only apply translational motion though (as far as I know), so the only force that would apply torque is friction, which wouldn't be specified.

(Also I asked my q in case you didn't see it)

Momentarily consider the ball to be a rod (same mechanics) and consider the pivot point to be the contact point between the rod and the ground. Friction can be neglected entirely, otherwise I do not believe that the above method of solution works because it neglects the work done by friction (sorry, should have specified in my original post). Because the normal force on the rod acts through the pivot point, the lever arm is zero and this force can be neglected. Thus, the only force acting upon the rod is the weight of the rod, but this weight acts as a torque because it is not acting through the pivot point, which causes the rod to tip over down the incline and fall. Applying this same principle to a ball, this torque would cause the ball to begin rotating.

the force that the carriage applies on the horse is equal to the force that the horse applies on the carriage, not the total force that the horse applies. The horse applies enough force on itself and the carriage to move them both but that force is not equal to the force that the horse applies on the carriage and the force that the carriage applies on the horse.

There are two components to this, the carriage moving and the horse moving. The equal and opposite forces described by Newton's Third Law are acting on different objects, the horse on the carriage and the horse on the carriage. In this case, the carriage only has one force on it, so it accelerates forward. However, the horse has two forces on it, the force of the carriage pulling it back and the force of the ground pushing it forward. As long as the force the horse applies on the ground (which is equal and opposite to the force the ground applies to it) is larger than the force the carriage applies to the horse, the horse moves.

Also, the pivot point of a ball rolling down a surface would be the ball's CoM (the ball rotates around its CoM), and both the force of gravity and the normal force act through the CoM, giving it zero torque. (Source: stackexchange)

EDIT: However, that does work if the ball is rolling without slipping which comes from static friction.

the force that the carriage applies on the horse is equal to the force that the horse applies on the carriage, not the total force that the horse applies. The horse applies enough force on itself and the carriage to move them both but that force is not equal to the force that the horse applies on the carriage and the force that the carriage applies on the horse.