Motion of an Insect in the Rough Bowl

Motion Of An Insect In the Rough Bowl

As the insect crawls up, limiting friction force decreases and the component of weight along the surface (driving force) will decrease. Let's assume the insect can crawl up to height 'h' before it starts slipping. At that moment the frictional force will be limiting friction force as shown in the figure.

Let m=mass of the insect, r=radius of the bowl, μ= coefficient of friction for limiting condition at point A.

R=mgcos⁡θ… (i) Limiting friction −fl=μR=μmgcos⁡θfll=mgsin⁡θ… (ii)

From equation (i) and (ii)-
mgsin⁡θ=μmgcos⁡θ
tan⁡θ=μtan2⁡θ=μ2
(r2−y2)y2=μ2y=r1+μ2r−h=r1+μ2 h=r[1+11+μ2]

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Till now we have studied the concept of Motion Of An Insect In The Rough Bowl. Now we will see some of the solved examples for a better understanding of the above concept.

Solved ExampleBased On Motion Of An Insect In The Rough Bowl

Example 1: An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle θ with the vertical, the maximum possible value of θ is given by:

1) cot−1⁡3
2) tan−1⁡3
3) sec−1⁡3
4) cosec−13

Solution:

As the insect crawls up, limiting friction force decreases and the component of weight along the surface (driving force) will decrease. Let's assume the insect can crawl upto angle $\theta$ before it starts slipping. At that moment the frictional force will be limiting friction force as shown in the figure.

From fig,

R=mgcos⁡θ Limiting friction −fl=μR=μmgcos⁡θ… For equilibrium- fl=mgsin⁡θ… (2) From equation (1) and (2)- mgcos⁡θ=μmgcos⁡θcot⁡θ=1μ=3θ=cot−1⁡3

Hence, the answer is option (1).

Example 2: If the coefficient of friction between on insect and bowl is μ and the radius of the bowl is r. Find the maximum height to which the insect can crawl up in the bowl.

1) r1+μ2
(2) rμ1+μ2
3) r[1−11+μ2]
4) ru1−μ2

Solution:

Motion an insect in the rough bowl -

Till the component of its weight along with the bowl is balanced by limiting frictional force, then only the insect crawls up the bowl, up to a certain height h

R=mgcos⁡θ......(I)
and
Fl=mgsin⁡θ ...... (ii)
Dividing (ii) by (i)
tan⁡θ=F1R=μ[ using Fl=μR]∴r2−y2y=μ or y=r1+μ2

So h=r−y=r[1−11+μ2],

∴h=r[1−11+μ2]

where,

h= height up to which insect can climb
m= mass of insect
r= radius of the bowl
μ= coefficient of friction

Example 3: An insect is at the bottom of a hemisphere ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height ' h ' from the bottom. If the coefficient of friction b/w the ground and the insect is 0.75. Then h is : (g=10 m/s2)

1) 0.6 m

2) 0.45 m

3) 0.2 m

4) 0.8 m

Solution:

Till the component of its weight along with the bowl is balanced by limiting frictional force, then only the insect crawls up the bowl, up to a certain height h

R=mgcos⁡θ
(i) and
Fl=mgsin⁡θ
Dividing (ii) by (i)
tan⁡θ=FFR=μ[ using Fl=μR]∴r2−y2y=μ or y=r1+μ2

So h=r−y=r[1−11+μ2],

∴h=r[1−11+μ2]

h=r[1−11+μ2]=1[1−11+(34)2]=1(1−45)=15=0.2 m

Hence, the correct answer is option (2).

Example 4: A block of mass m is placed on a surface with a vertical cross-
section given by y=x36. If the coefficient of friction is 0.5 ,t he maximum height above the ground at which the block can be placed without slipping is :

1) 16m
2) 23m
3) 13m
4) 12m

Solution:

As we learnt in

mgSinθ=μmgCosθtan⁡θ=μ⇒>y=x36dydx=x22=12x=1y=16m

Hence, the answer is the option (1).

Summary

In Short, the motion of an insect in a rough bowl is influenced by factors such as friction, gravity, and the bowl's curvature. The insect's path is determined by the balance between its own movement and the forces acting on it, often resulting in complex, non-linear trajectories. Understanding these dynamics can provide insights into the insect's behaviour and the physical principles at play.