English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

The angular position of a point on the rim of a rotating wheel is given by θ = 4.24t - 2.17t2 + 3.51t3, where θ is in radians and t is in seconds. What are the angular velocities at (a) t = 1.09 s and (b) t = 6.17 s? (c) What is the average angular acceleration for the time interval that begins at t = 1.09 s and ends at t = 6.17 s? What are the instantaneous angular accelerations at (d) the beginning and (e) the end of this time interval?

2007-12-05 08:36:35 · 3 answers · asked by shade o 1 in Science & Mathematics Physics

3 answers

ω=dθ/dt
and
α=dω/dt

solve for ω
ω=4.24-4.34*t+10.53*t^2
a)
ω(1.09)=4.24-4.34*1.09+10.53*1.09^2
=12.02 rad/s
b)
ω(6.17)=4.24-4.34*6.17+10.53*6.17^2
=378.33 rad/s

c)
=(378.33-12.02)/(6.17-1.09)
72.108 is the average α

d)
α=-4.34+21.06*t

α(1.09)=-4.34+21.06*1.09
18.62 rad/s^2
e)
α(6.17)=-4.34+21.06*6.17
125.6 rad/s^2

j

2007-12-05 13:05:33 · answer #1 · answered by odu83 7 · 0 0

The given relation is -
θ = 4.24 t - 2.17 t^2 + 3.51 t^3
Differentiating w.r.to t we get the expression for the angular velocity.

dθ/dt = 4.24 - 2 ( 2.17 ) t + 3 ( 3.51) t^2 = w (say)
Then w = 10.53 t^2 - 4.34 t + 4.24 radians / sec

Hence -

w at t (= 1.09 s) = 10. 53 (1.09)^2 - 4.34 (1.09) + 4.24 = 12.02 Rad/s .............. Answer (a)

(b) w (t = 6.17) = 10. 53 (6.17)^2 - 4.34 (6.17) + 4.24 = 378.33 Rad / sec ..................... Answer (b)

(c) Change in velocity during the specified time interval (6.17 - 1.09 s ie in 5.08 s )

Change in velocity = ( 378.33 - 12.02 ) = 366.31 Rad/s

Hence Average Angular accn (alpha) = w / time duration
= 366.31 / 5.08 = 72. 108 Rad / sec^2 ......... Answer

Again as angular velocity w = 10.53 t^2 - 4.34 t + 4.24 radians / sec,
Differentiating again w.r.to t we get the instantaneous accn =
21.06 t - 4.34 Rad/s/s


Instantaneous accn (at t = 1.09 Sec.) = 21.06 (1.09) - 4.34 = 19.204 Rad/s/s . ........ Answer

Instantaneous accn (at t = 6.17 Sec.) = 21.06 (6.17) - 4.34 = 125.6 Rad/s/s . ........ Answer

2007-12-05 13:08:06 · answer #2 · answered by Pramod Kumar 7 · 0 0

We know that
ω=dθ/dt
and
α=dω/dt

solve for ω
ω=4.24-4.34*t+10.53*t^2
a)
ω(1.09)=4.24-4.34*1.09+10.53*1.09^2
=12.02 rad/s
b)
ω(6.17)=4.24-4.34*6.17+10.53*6.17^2
=378.33 rad/s

c)
=(378.33-12.02)/(6.17-1.09)
72.108 is the average α

d)
α=-4.34+21.06*t

α(1.09)=-4.34+21.06*1.09
18.62 rad/s^2
e)
α(6.17)=-4.34+21.06*6.17
125.6 rad/s^2

2007-12-05 18:07:42 · answer #3 · answered by kartheek 2 · 0 0

fedest.com, questions and answers