# Angular Velocity and other problems

• Apr 11th 2009, 05:54 AM
taichi2910
Angular Velocity and other problems
1) A rotating body has an angular displacement of (teta) radians in time (t) seconds, the relationship between displacement and the time is

(teta) = 7t^3-5t^2+30

find:
angulaer velocity when t = 3second
the acceleration when t = 3
the time when the acceleration = 0

2)
water drain from a cylindrical tank of radius 4 meter. as the height h falls the pressure reduces and it is observer that the relationship h and time t is given by
h= 5e^t/21

find the rate that the heiht is falling after 3 s
the flow rate in liter persecond after 3 seconds

please can someone help me in solving the above problems and an explaination on the method .

• Apr 11th 2009, 06:23 AM
Differentiating
Hello taichi2910
Quote:

Originally Posted by taichi2910
1) A rotating body has an angular displacement of (teta) radians in time (t) seconds, the relationship between displacement and the time is

(teta) = 7t^3-5t^2+30

find:
angulaer velocity when t = 3second
the acceleration when t = 3
the time when the acceleration = 0

You are told:

$\theta = 7t^3 - 5t^2 + 30$

and you are asked to find

• The value of the angular velocity - that's $\frac{d\theta}{dt}$ - when $t = 3$. So differentiate and plug in the value $t = 3$.

• The value of the angular acceleration - that's $\frac{d^2\theta}{dt^2}$ - when $t = 3$. Differentiate again, and put $t = 3$.

• The value of $t$ when $\frac{d^2\theta}{dt^2} = 0$. So use your equation in (2) and find the value of $t$ that will make the expression zero.

Can you do this now?

Quote:

water drain from a cylindrical tank of radius 4 meter. as the height h falls the pressure reduces and it is observer that the relationship h and time t is given by
h= 5e^t/21

find the rate that the heiht is falling after 3 s
the flow rate in liter persecond after 3 seconds
You have given us the equation $h = \frac{5e^t}{21}$. Are you sure this is correct? Should it be $h = 5e^{-\frac{t}{21}}$. This would certainly make more sense.

Whatever the formula, you need the rate that the height is falling - that's $\frac{dh}{dt}$ - when $t = 3$ (and it should be negative - hence I suspect you have missed a minus sign).

So differentiate and put $t = 3$. This will give you the rate at which the height is falling in $ms^{-1}$.

Then you need to convert this answer into a volume flow in l/sec. So find the cross-sectional area of the tank ( $A = \pi r^2$), and multiply your first answer by this area. This will give the rate in $m^3s^{-1}$. There are 1000 l in 1 $m^3$. So divide by 1000.

• Apr 11th 2009, 07:01 AM
taichi2910
I am really bad when it comes to physics and maths, i cant remember how to solve any of this problems even with your explanation which is clears. left high school 12 years ago and i have forgotten almost everything i did in maths and physics (really bad) then. just stated college and the works are scary. so please can you help me a bit with the solving?

• Apr 11th 2009, 08:57 AM
Hello taichi2910

Quote:

You are told:

$\theta = 7t^3 - 5t^2 + 30$

and you are asked to find

• The value of the angular velocity - that's $\frac{d\theta}{dt}$ - when $t = 3$. So differentiate and plug in the value $t = 3$.

$\color{red}\frac{d\theta}{dt} = 21t^2-10t$

• The value of the angular acceleration - that's $\frac{d^2\theta}{dt^2}$ - when $t = 3$. Differentiate again, and put $t = 3$.

$\color{red}\frac{d^2\theta}{dt^2}=42t-10$

• The value of $t$ when $\frac{d^2\theta}{dt^2} = 0$. So use your equation in (2) and find the value of $t$ that will make the expression zero.

$\color{red}42t-10=0$

Can you do this now?

You have given us the equation $h = \frac{5e^t}{21}$. Are you sure this is correct? Should it be $h = 5e^{-\frac{t}{21}}$. This would certainly make more sense.

Whatever the formula, you need the rate that the height is falling - that's $\frac{dh}{dt}$ - when $t = 3$ (and it should be negative - hence I suspect you have missed a minus sign).

So differentiate and put $t = 3$. This will give you the rate at which the height is falling in $ms^{-1}$.

Assuming the expression is as I supposed:

$\color{red}\frac{dh}{dt}= -\frac{5}{21}e^{-\frac{t}{21}}$

Then you need to convert this answer into a volume flow in l/sec. So find the cross-sectional area of the tank ( $A = \pi r^2$), and multiply your first answer by this area. This will give the rate in $m^3s^{-1}$. There are 1000 l in 1 $m^3$. So divide by 1000.