42
1 Kinematics and Statics
Thus the CM is located at a height h −
3
4 h =
1
4 h above the centre of the base
of the cone.
1.53 Take the origin at O, Fig. 1.32. Let the mass of the wire be M. Consider mass
element dm at angles θ and θ + dθ
Fig. 1.32
dm =
M R dθ
2α R
=
M dθ
2α
(1)
From symmetry the CM of the wire must be on the y-axis.
The y-coordinate of dm is y = R sin θ
y CM =
1
M
y dm =
90+α
90−α
R sin θ dθ
2α
=
R sin α
α
Note that the results of prob. (1.43) follow for α =
1
2 π .
1.54 V CM =
m i v i
m i
=
4mv 0 + (m)(0)
5m
=
4v 0
5
1.55
ρ = cx(c = constant); dm = ρ dx = cx dx
x CM =
x dm
dm
=
L
0 xcx dx
L
0 cx dx
=
2
3
L
1.56 x CM =
m i x i
m i
=
m L + (2m)(2L) + (3m)(3L) + · · · + (nm)(nL)
m + 2m + 3m + · · · + nm
=
(1 + 4 + 9 + · · · + n 2 )L
1 + 2 + 3 + · · · + n
=
(sum of squares of natural numbers)L
sum of natural numbers
=
n(n + 1)(2n + 1)L/6
n(n + 1)/2
= (2n + 1)
L
3