SRL Archive - Science

*note* tension on either side of the pulley is not the same.

Brain wanna answer this for me? :p

is the answer acceleration a = 2.60 m/s^2 ?
(normally they give you the numerial answer in questions like this)

uh geez…..physics, let me see what I can do.


edit:
brb like 5 mins

Edit: alright, uh…any way I need to solve it? work/energy, kinematics?

bah I don't know, looked through my dynamics notes and homework and couldn't find anything, should be pretty straight forward though. What book are you guys using?

From: Yakman
is the answer acceleration a = 2.60 m/s^2 ?
(normally they give you the numerial answer in questions like this)

ill check…i have 5 tries to get the answer right.

@Brain. we're using kinematics equations.

edit: yakman you got the answer. care explaining?

you doing web assign? What book? :p

web assign? its a capa assignment. our text is Physics for scientists and engineers by Serway and Jewett

I have that book, lol :p almost done with it too

haha nice!

right, I'll be explaining.

In this problem the mass of the pulley is not constant, so we need to take its acceleration into account.


I've renamed the motor "the engine", because otherwise both of those things start with m.

Draw the free body force diagrams.

Tension from the engine and Tension from the mass.


goto www.mathbin.net, copypaste this LaTeX in and press preview to see my maths.

Finding the resultant force and applying N2
[ieq]
\sum F_m = T - mg = ma

T = ma + mg
[/ieq]

Finding the resultant torque and applying N2
[ieq]\sum \tau_p = rT_e - rT = I\alpha[/ieq]

Using
[ieq]\alpha = \frac{a}{r}[/ieq] Oh btw, the acceleration of the mass is equal to the acceleration of the pulley

Solve for acceleration a
[ieq]rT_e - rT = I\frac{a}{r}[/ieq]
[ieq]r^2T_e - r^2(ma + mg) = Ia[/ieq]
[ieq]r^2T_e - r^2mg = Ia + r^2ma[/ieq]
[ieq]r^2T_e - r^2mg = a(I + r^2m)[/ieq]
[ieq]a = \frac {r^2T_e - r^2mg} {I + r^2m}[/ieq]

now substitute all your values

[ieq]a = \frac {0.725^2 \times 1.28 \times 10^4 - 0.725^2 \times 1000 \times 9.81} {78.5 + 0.725^2 \times 1000} = 2.60 m s^-2[/ieq]

Thanks a lot Yakman! makes perfect sense now.