Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 3 (Q.No. 21)
21.
A stone of mass 1 kg is tied to a string of length 1 m and whirled in a horizontal circle at a constant angular speed 5 rad/sec. The tension in the string is,
Discussion:
4 comments Page 1 of 1.
Anc said:
7 years ago
A one kg mass tied at the end of the string 0.5 m long is whirled in a horizontal circle, the other end of the string being fixed. The breaking tension in the string is 50 N. Find the greatest speed that can be given to the mass.
Solution:
Given: Mass of a body = m = 1 kg, radius of circular path = r = 0.5 m, Centripetal force = F = 50 N.
To find: Greatest speed = v = ?,
The necessary centripetal force acting on a body is given by tension in the string.
Ans: The greatest speed can be given to mass = 5 m/s.
Solution:
Given: Mass of a body = m = 1 kg, radius of circular path = r = 0.5 m, Centripetal force = F = 50 N.
To find: Greatest speed = v = ?,
The necessary centripetal force acting on a body is given by tension in the string.
Ans: The greatest speed can be given to mass = 5 m/s.
Ihsan Ullah said:
5 years ago
The tension in the string will be equal to the centrifugal force, so,
T = Fc.
T = (mv^2)/r or T=mw^2r As w=v/r.
So put the values of each,
T = (1)*(5^2)*(1),
T = 25N.
T = Fc.
T = (mv^2)/r or T=mw^2r As w=v/r.
So put the values of each,
T = (1)*(5^2)*(1),
T = 25N.
(2)
Bret said:
9 years ago
@Rakesh.
Is it correct?
Is it correct?
Rakesh Dabhi said:
1 decade ago
mv^2/r = (1*25)/1 = 25.
(1)
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