Solved it Nope. - Piano slipping down a ramp

[Medgirl314]

Solved it! Nope. -- Piano slipping down a ramp

Homework Statement

Two men loading a 900 kg piano onto a truck lose control and the piano slips, rolling down the loading ramp. The ramp is a 12 degree incline that is 3.2 m long. How long does it take the piano to reach the bottom of the ramp? Assume there is no friction.

Homework Equations

My physics teacher found this equation for acceleration for an object on an incline:
a=g(sin)(theta)-(mu)g(cos)(theta)

I tried tweaking it a bit. I think the (mu)g(cos) can be left out since mu is zero.

a=g(sin)theta
a=9.8(.20)
a=2.03 m/s^2

Easy, right?

The Attempt at a Solution

Nope. Not easy. The equation I found above doesn't take mass into consideration. Using this acceleration as my change in velocity, I get t=2.03/3.2=0.63 s. Reasonable enough. But what about the mass? Where did I go wrong?

Thanks in advance!

 

[BvU]

Not good. Acceleration is change in velocity. You've learned other expressions for linear motion with constant acceleration. List a few under 2..

And: acceleration/length is not the dimension of time !
Also: $\sin \left( \frac {12}{180}\pi\right) = 0.2$ is a bit coarse

 

[Rococo]

Homework Statement

Two men loading a 900 kg piano onto a truck lose control and the piano slips, rolling down the loading ramp. The ramp is a 12 degree incline that is 3.2 m long. How long does it take the piano to reach the bottom of the ramp? Assume there is no friction.

Homework Equations

My physics teacher found this equation for acceleration for an object on an incline:
a=g(sin)(theta)-(mu)g(cos)(theta)

I tried tweaking it a bit. I think the (mu)g(cos) can be left out since mu is zero.

a=g(sin)theta
a=9.8(.20)
a=2.03 m/s^2

Easy, right?

The Attempt at a Solution

Nope. Not easy. The equation I found above doesn't take mass into consideration. Using this acceleration as my change in velocity, I get t=2.03/3.2=0.63 s. Reasonable enough. But what about the mass? Where did I go wrong?

Thanks in advance!

Your acceleration is correct.

Your value for the time taken is incorrect.

You know that $a=2.03 ms^{-2}$. You know that it travels a distance $s=3.2 m$. We also know that the piano's initial speed is zero, so $u=0 ms^{-2}$.

You want to find the time taken, $t$.

Do you know a formula that connects $s$, $u$, $a$ and $t$?

 

[Medgirl314]

Yeah, I didn't think so, but there have been certain instances where the acceleration was the velocity, so I figured there was no harm in playing with it.

a=Δv/Δt
x=x0+v0t+1/2at^2
v^2=v0^2+2aΔx

Hm. Would the last one work? If so, why don't I need to account for the mass?

 

[Medgirl314]

Your acceleration is correct.

Your value for the time taken is incorrect.

You know that $a=2.03 ms^{-2}$. You know that it travels a distance $s=3.2 m$. We also know that the piano's initial speed is zero, so $u=0 ms^{-2}$.

You want to find the time taken, $t$.

Do you know a formula that connects $s$, $u$, $a$ and $t$?

What's $u$ ? I think we may have different variables for that quantity.

 

[Rococo]

Yeah, I didn't think so, but there have been certain instances where the acceleration was the velocity, so I figured there was no harm in playing with it.

a=Δv/Δt
x=x0+v0t+1/2at^2
v^2=v0^2+2aΔx

Hm. Would the last one work? If so, why don't I need to account for the mass?

You are trying to find the time taken, $t$. Therefore the equation you choose must have $t$ in it.

The middle equation is the one you need. Using the symbols I used in my previous post, the equation can be written:

$s=ut + \frac{1}{2}at^{2}$

If you were to place a golf ball and a bowling ball at the top of the ramp, they would both roll down, and reach the bottom of the ramp at exactly the same time. This might seem surprising, but the time taken for an object to roll down the ramp does not depend on the mass of the object. The acceleration for any object will be the same, $a=gsinθ$ , as you correctly showed.

 

[Rococo]

What's $u$ ? I think we may have different variables for that quantity.

Yes, I usually use the symbol $u$ to mean the initial velocity of the object. It is the same thing as $v_0$.

 

[Medgirl314]

Sorry, I mis-stated my question about the mass. I understand that the time doesn't depend on the mass, but I was wondering why it was included in this problem. Just to throw me off? It worked. ;-)

s=ut+1/2 at^2
s=1/2at^2

Major brain lapse. When we say at^2, do we mean just the t is squared?

Thanks again!

 

[Rococo]

Sorry, I mis-stated my question about the mass. I understand that the time doesn't depend on the mass, but I was wondering why it was included in this problem. Just to throw me off? It worked. ;-)

s=ut+1/2 at^2
s=1/2at^2

Major brain lapse. When we say at^2, do we mean just the t is squared?

Thanks again!

Yes, the mass of piano is not needed to solve the problem.

Yes, just the t is squared: $\frac{1}{2}at^2=\frac{1}{2} \times a \times t \times t$

So, you have the correct equation:

$s=\frac{1}{2}at^2$

You know that $s=3.2m$ and $a=2.03 ms^{-2}$.

Can you now use the equation to find $t$?

 

[Medgirl314]

s/a=1/2t^2
3.2/2.03=1/2t^2
1.57=1/2t^2
3.14=t^2
1.77=t.

Okay, that may not be right. But I really, really, really want it to be because pi.

 

[Rococo]

s/a=1/2t^2
3.2/2.03=1/2t^2
1.57=1/2t^2
3.14=t^2
1.77=t.

Okay, that may not be right. But I really, really, really want it to be because pi.

Yes, that's correct.

 

[Medgirl314]

Yay! My physics teacher is epic. Thank you!

 

[BvU]

See, some physicists are not too bad... almost human