Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 30)
30.
The piston of a steam engine moves with a simple harmonic motion. The crank rotates 120 r.p.m. and the stroke length is 2 metres. The linear velocity of the piston when it is at a distance of 0.5 metre from the centre, is
Discussion:
6 comments Page 1 of 1.
Abhay said:
10 years ago
Given: N = 120 R.P.M. or ω = 2*120/60 = 4 rad/s ; 2r = 2 m or r = 1 m;
x = 0.5 m.
Velocity of the piston. We know that velocity of the piston,
Linear velocity, v = ω r^2/x^2 2^2/(0.5)^2 = 10.88 m/s Answer.
x = 0.5 m.
Velocity of the piston. We know that velocity of the piston,
Linear velocity, v = ω r^2/x^2 2^2/(0.5)^2 = 10.88 m/s Answer.
Geo said:
9 years ago
Please explain the equation.
Harshit Dharsandia said:
8 years ago
Here, w= (2 * 3.14 * N)/60.
& Velocity v= w(root(r^2-x^2)).
& Velocity v= w(root(r^2-x^2)).
Amit said:
8 years ago
Please explain the answer with the correct formulas.
AKR said:
8 years ago
w = 2 * Pi * N/60 = 2 * 3.14 * 120/60 = 12.56 radiance,
v= w * sqrt (r^2-x^2),
v =12.56 * sqrt (1-.25) .....x= 0.5, r=1.
v = 10.88 m/s.
v= w * sqrt (r^2-x^2),
v =12.56 * sqrt (1-.25) .....x= 0.5, r=1.
v = 10.88 m/s.
(1)
Arpit said:
7 years ago
X=Asinwt
V=Awcoswt
Here, A=1m(half of stroke), X=.5.
Put the values and find v.
V=Awcoswt
Here, A=1m(half of stroke), X=.5.
Put the values and find v.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers