# Solving Rotational Motion of Wheel with 3.21 kg Block

• seraphimhouse
In summary, a wheel of radius 0.358 m with a rotational inertia of 0.0421 kg·m2 is connected to a 3.21 kg block via a massless cord. When a horizontal force of magnitude P = 4.31 N is applied to the block, the wheel experiences an angular acceleration of 4.18 rad/s^2 in the clockwise direction. By setting the tension in the cord equal to the net torque and solving for acceleration, we can find the correct value of -1.496 m/s^2. The computer may have been thrown off by a missing minus sign in the calculations.
seraphimhouse

## Homework Statement

A wheel of radius 0.358 m is mounted on a frictionless horizontal axis. The rotational inertia of the wheel about the axis is 0.0421 kg·m2. A massless cord wrapped around the wheel is attached to a 3.21 kg block that slides on a horizontal frictionless surface. If a horizontal force of magnitude P = 4.31 N is applied to the block as shown in Fig. 10-54, what is the angular acceleration of the wheel? Take the clockwise direction to be the negative direction and assume the string does not slip on the wheel.

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c10/fig10_54.gif

## Homework Equations

tnet = I(alpha)

at [tangential acceleration] = (alpha)r

## The Attempt at a Solution

I did the forces of the box and got that Ft = p [the horizontal force] - ma where m is the mass of the box.
So, Ft = p - ma

I found the net torque of the disk rotating at the center of its axis to be -Ftr and the rotational inertia of the disk to be I = 1/2Mr^2 where M is the mass of the disk.
So, Tnet = -Ftr & I = 1/2Mr^2

We know that Tnet = I[alpha] so therefore,
-Ftr = 1/2Mr^2[alpha]

We know that the string does not slip on the wheel, therefore the tangential acceleration is equal to the angular acceleration times it's radius.
at = [alpha]r
[alpha] = at/r

So by substituting it to the previous equation, I get
-Ftr = 1/2Mr^2(at/r)
Cleaning up this equation I get
Ft = -1/2Ma

Before I go any further, I found the mass of the disk by using I = 1/2Mr^2. We know that I=0.0421 kg m^2.
0.0421 = 1/2Mr^2 we know what r is so therefore,
M = .657 kg

So I then set the tension found using the disk and the free body diagram of the box equal to each other to get
P - ma = -1/2Ma
cleaning this up and having acceleration set on one side of the equation I get,
a = 2p/[2m-M]
a = 1.496 m/s^2

And finally I substituted acceleration back into the tangential acceleration equation which was
[alpha] = a/r

Probably a mistake to assume I = .5mr².
You can figure out the rotational part from
torque = Iα
Tr = I*a/r (where T = tension)
without the assumption.
If you sub this expression for T into the T = p - ma you can solve for acceleration and you'll get a smaller value.

I substituted it for T = I$$\alpha$$
I used -Ftr = I [a/r]
Them subbed in Ft = P - ma
rma - rP = I a/r

I cleaned it up and still got 4 m/s^2 which still gave me 20 rad/s^2

I used -Ftr = I [a/r]
Ah, I didn't have that minus sign. Must rethink the whole thing!
The question specifies counterclockwise as positive, so the first equation should be Ft - p = ma
And the circular one Ft*r = I*a/r
So I*a/r² - p = ma
(I/r² - m) a = p
a = p/(I/r² - m)
a = 4.31/(.0421/.358² - 3.21) = -1.496
so we agree on the number except for the sign. Could the computer be upset over the missing minus sign? It clearly is correct - the p force makes it accelerate clockwise which is defined as negative.

## 1. How do you calculate the rotational motion of a wheel with a 3.21 kg block?

To calculate the rotational motion of a wheel with a 3.21 kg block, you can use the equation: τ = Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. You will also need to know the radius of the wheel and the force applied to the block.

## 2. What is the moment of inertia?

The moment of inertia is a measure of an object's resistance to changes in its rotation. It is calculated by multiplying the mass of an object by the square of its distance from the axis of rotation.

## 3. How does the mass of the block affect the rotational motion of the wheel?

The mass of the block affects the rotational motion of the wheel by changing the moment of inertia. A larger mass will result in a larger moment of inertia, which will require more torque to achieve the same angular acceleration.

## 4. Can you solve for the rotational motion without knowing the mass of the block?

No, the mass of the block is a necessary component in calculating the rotational motion. Without knowing the mass, you cannot accurately calculate the moment of inertia and therefore cannot solve for the rotational motion.

## 5. How does friction impact the rotational motion of the wheel and block?

Friction can affect the rotational motion of the wheel and block by creating a torque that opposes the motion. This can cause the wheel to slow down or even stop rotating. To account for friction, you can include it as a negative torque in the equation τ = Iα.

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