Determine energy required to stop rolling mass

[I_Try_Math]

Homework Statement

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?

Relevant Equations

$K_T = \frac{1}{2}mv^2$
$K_R = \frac{1}{2}Iw^2$

Is it possible to solve this without knowing the radius of the cylinder? My initial thoughts were that the energy required to stop it would be the sum of its rotational and translation kinetic energy, but I'm not sure it can be calculated without knowing the radius.

 

[erobz]

Homework Statement: A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?
Relevant Equations: $K_T = \frac{1}{2}mv^2$
$K_R = \frac{1}{2}Iw^2$

Is it possible to solve this without knowing the radius of the cylinder? My initial thoughts were that the energy required to stop it would be the sum of its rotational and translation kinetic energy, but I'm not sure it can be calculated without knowing the radius.

It's implied that its rolling without slipping. Show an attempt at this computation.

 

[I_Try_Math]

It's implied that its rolling without slipping. Show an attempt at this computation.

$\frac{1}{2}mv^2 + \frac{1}{2}Iw^2 =$
$\frac{1}{2}(40 kg)(6 m/s)^2 + \frac{1}{2}(\frac{1}{2}(40 kg)r^2)(\frac{6 m/s}{r})^2 =$
Ah right, if my math is correct the radii cancel out. Thanks for the hints @erobz.

 

[erobz]

$\frac{1}{2}mv^2 + \frac{1}{2}Iw^2 =$
$\frac{1}{2}(40 kg)(6 m/s)^2 + \frac{1}{2}(\frac{1}{2}(40 kg)r^2)(\frac{6 m/s}{r})^2 =$
Ah right, if my math is correct the radii cancel out. Thanks for the hints @erobz.

You're welcome!

Future note: It's better to work in all variables until the last step.

$$ = \frac{3}{4}Mv^2 $$

 

[erobz]

Also... I think if we are being precise what should the sign be on the work done on the cylinder to stop it?

 

[I_Try_Math]

Also... I think if we are being precise what should the sign be on the work done on the cylinder to stop it?

True, I shouldn't say the energy required to stop the mass is equal to the mass's kinetic energy. It's the negative of the mass's kinetic energy.

 

[erobz]

True, I shouldn't say the energy required to stop the mass is equal to the mass's kinetic energy. It's the negative of the mass's kinetic energy.

I know it seems trivial, but I also know from personal experience that hastily talking about the just the absolute values can get you in trouble when things get a bit more convoluted.

Have a good one!

 

[haruspex]

I'm not sure it can be calculated without knowing the radius.

Are you sure it cannot?

 

[I_Try_Math]

I know it seems trivial, but I also know from personal experience that hastily talking about the just the absolute values can get you in trouble when things get a bit more convoluted.

Have a good one!

For sure, I've certainly had my fair share of frustrations only to find out they were due to a sign error. Good to keep in mind!

 

[I_Try_Math]

Are you sure it cannot?

Now that I've done the calculation I can see that isn't necessary to know the radius. It still seems slightly counterintuitive in my opinion but the math doesn't lie, so to speak.

 

[haruspex]

Now that I've done the calculation I can see that isn't necessary to know the radius. It still seems slightly counterintuitive in my opinion but the math doesn't lie, so to speak.

It can be deduced by dimensional analysis. It is evident that the KE can only depend on the speed, mass and radius. In dimensional terms:
$[KE]=[v]^a[m]^b[r]^c$
$ML^2T^{-2}=(LT^{-1})^aM^bL^c$
Whence a=2, b=1, c=0.