A uniform solid cone of height h = 5 m and base radius 3m is moving on a smooth horizontal table. The velocity of centre of mass of cone is 'V' in horizontal direction and cone rotates about its vertical axis with angular velocity a = 10rad/s. Then find the minimum velocity v (in m/s) to avoid bumping the edge of the table when it gets there. (Take g = 10 m/s2) 1
Answers
The moment of inertia of the cone around the axis depicted in the illustration is = 10 3.
mR
2
Hence, R=radius of cone
m=mass of cone
When the particle is at apex it has no moment of inertia about axis of rotation .
When the particle is at the base of cone it has a moment of inertia which is equal to m/s
2
.
So, initially the moment of inertia of system is
I
i
=
10
3
mR
2
+0
=
10
3
mR
2
Finally the moment of inertia of the system
I
f
=
10
3
mR
2
+mR
2
=
10
13
mR
2
So according to conservation of angular momentum
I
i
ω=I
f
ω
′
⟹ω
′
=
I
f
I
i
′
ω
ω
′
=
(
10
13
)mR
2The moment of inertia of the cone around the axis depicted in the illustration is = 10 3.
mR
2
Hence, R=radius of cone
m=mass of cone
When the particle is at apex it has no moment of inertia about axis of rotation .
When the particle is at the base of cone it has a moment of inertia which is equal to m/s
2
.
So, initially the moment of inertia of system is
I
i
=
10
3
mR
2
+0
=
10
3
mR
2
Finally the moment of inertia of the system
I
f
=
10
3
mR
2
+mR
2
=
10
13
mR
2
So according to conservation of angular momentum
I
i
ω=I
f
ω
′
⟹ω
′
=
I
f
I
i
′
ω
ω
′
=
(
10
13
)mR
2The moment of inertia of the cone around the axis depicted in the illustration is = 10 3.
mR
2
Hence, R=radius of cone
m=mass of cone
When the particle is at apex it has no moment of inertia about axis of rotation .
When the particle is at the base of cone it has a moment of inertia which is equal to m/s
2
.
So, initially the moment of inertia of system is
I
i
=
10
3
mR
2
+0
=
10
3
mR
2
Finally the moment of inertia of the system
I
f
=
10
3
mR
2
+mR
2
=
10
13
mR
2
So according to conservation of angular momentum
I
i
ω=I
f
ω
′
⟹ω
′
=
I
f
I
i
′
ω
ω
′
=
(
10
13
)mR
2The moment of inertia of the cone around the axis depicted in the illustration is = 10 3.
mR
2
As a result, R=cone radius
m is the cone's mass.
When the particle is at apex it has no moment of inertia about axis of rotation .
When the particle is at the base of cone it has a moment of inertia which is equal to m/s
2
.
So, initially the moment of inertia of system is
I
i
=
10
3
mR
2
+0
=
10
3
mR
2
Finally the moment of inertia of the system
I
f
=
10
3
mR
2
+mR
2
=
10
13
mR
2
So according to conservation of angular momentum
I
i
ω=I
f
ω
′
⟹ω
′
=
I
f
I
i
′
ω
ω
′
=
(
10
13
)mR
2The moment of inertia of the cone around the axis depicted in the illustration is = 10 3.
mR
2
As a result, R=cone radius
m is the cone's mass.
When the particle is at apex it has no moment of inertia about axis of rotation .
When the particle is at the base of cone it has a moment of inertia which is equal to m/s
2
.
So, initially the moment of inertia of system is
I
i
=
10
3
mR
2
+0
=
10
3
mR
2
Finally the moment of inertia of the system
I
f
=
10
3
mR
2
+mR
2
=
10
13
mR
2
So according to conservation of angular momentum
I
i
ω=I
f
ω
′
⟹ω
′
=
I
f
I
i
′
ω
ω
′
=
(
10
13
)mR
2