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(a) What is the tangential acceleration of a bug on therim of a 8.0 in. diameter

ID: 1728399 • Letter: #

Question

(a) What is the tangential acceleration of a bug on therim of a 8.0 in. diameter disk if the diskmoves from rest to an angular speed of 78revolutions per minute in 5.0 s?
1m/s2
(b) When the disk is at its final speed, what is the tangentialvelocity of the bug?
2 m/s
(c) One second after the bug starts from rest, what is itstangential acceleration?
3m/s2
What is its centripetal acceleration?
4m/s2
What is its total acceleration?
5m/s2
6° (relativeto the tangential acceleration)
N/b Please no partial solutions for it will not halpme. (a) What is the tangential acceleration of a bug on therim of a 8.0 in. diameter disk if the diskmoves from rest to an angular speed of 78revolutions per minute in 5.0 s?
1m/s2
(b) When the disk is at its final speed, what is the tangentialvelocity of the bug?
2 m/s
(c) One second after the bug starts from rest, what is itstangential acceleration?
3m/s2
What is its centripetal acceleration?
4m/s2
What is its total acceleration?
5m/s2
6° (relativeto the tangential acceleration)
N/b Please no partial solutions for it will not halpme.

Explanation / Answer


   from the theory we can see that
   1 rev = 2 rad and
   1 min = 60 s
   1 inch = 2.54 cm
   the final angular velocity will be
   f = 78 rev / min
        = ...... rad / s
   radius of the disk is
   r = 4.0 in
      = ..... cm
   t = 5.0 s
(a)
   the tangential acceleration of a bug as the diskspeeds up will be
   at = r
       = r ( /t)
       = ........ m /s2
(b)
   the final tangential speed of the bug will be
   vt = r f
       = ..... m / s
(c)
   at t = 1.0 s
   = i + t
       = 0 + ( t)
       = ........ rad / s
   thus the tangential acceleration will be
   at = r
   radial acceleration will be
   ac = r 2
       = ....... m /s2
   the total acceleration will be
   a = (ac2 +at2)
       = ....... m /s2
   the direction will be
   = tan-1(at /ac)
      = ..........o