By Team Aakash Byju's | 18th December 2022
A football of radius R is kept on a hole of radius r (r<R) made on a plank kept horizontally. One end of the plank is lifted so that it gets tilted making an angle θ from the horizontal as shown.
A.
B.
C.
D.
sinθ =
r
R
sinθ =
r
2R
cosθ =
r
R
tanθ =
r
R
Detailed Explanation
r
R
.
When the plank is lifted on one side making an angle θ from the horizontal, the normal reaction (N & N ) will be acting upon the point of contact.
1
2
θ
For the maximum value of θ, the ball will be about to roll. In such a case, the weight of the ball will be acting along the normal reaction.
θ
θ
max
max
r
Normal here tends to zero on verge of rolling
The torque balancing equation for the component of weight will be
θ
θ
max
max
r
Normal here tends to zero on verge of rolling
mg sinθ
R - r = mg cosθ(r)
2
√
2
=>
sinθ
cosθ
=
r
√
R - r
2
2
=>
tanθ =
r
√
2
2
R - r
√
2
R - r
2
Now, consider a right triangle to define the tan θ which is the ratio of the opposite and base.
r
θ
=>
tanθ =
r
2
√
2
R - r
√
2
R - r
2
Using the Pythagorean Theorem, let us find the hypotenuse as
r
θ
=
hyp
2
√
( )
2
R - r
hyp
2
2
r
+
2
=
2
r
+
2
R - r
2
hyp
2
=
2
R
hyp
=
R
=>
. .
.
sinθ =
opp
hyp
=
r
R