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Math Help - Angular Velocity and other problems

  1. #1
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    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 .

    thank you in advance.
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  2. #2
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    Differentiating

    Hello taichi2910
    Quote Originally Posted by taichi2910 View Post
    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?

    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.

    Grandad
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  3. #3
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    hello Grandad,
    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?

    thanks in advance.
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  4. #4
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    Hello taichi2910

    Quote Originally Posted by Grandad View Post
    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.

    Grandad
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