This is a classic and highly conceptual problem from JEE Advanced 2017 that tests your deep understanding of rigid body dynamics, torque, and the subtle nuances of mathematical language in physics problems. Let's break down the mechanics of a wheel climbing a step, analyzing each option carefully.
Analyzing the Setup
Imagine a wheel of radius R resting on the ground, touching a step that is also of height R. The corner of the step is our pivot point, Q. For the wheel to climb the step, it must rotate about this pivot point Q.
As the wheel climbs, its center C traces a circular arc around Q. We can track this motion using an angle θ, which is the angle the line QC makes with the horizontal. Initially, when the wheel is on the ground, θ=0∘. As it climbs, θ increases towards 90∘.
Option A
The Zero Torque
Let's evaluate the first option. A force is applied normal to the circumference at the leftmost point P. A force normal to a circle's circumference always points directly towards its center, C.
Because the step height is R and the wheel radius is R, the points P, C, and Q all lie on the exact same horizontal line initially. As the force points from P to C, its line of action extends straight through the pivot point Q.
Since the perpendicular distance from the pivot Q to the line of action of the force is zero, the torque of this applied force about Q is perfectly zero. Therefore, option (a) is correct.
Option B
The Tangential Force at S
What if we apply a tangential force at the topmost point S? A tangential force at the top of the wheel acts horizontally.
The perpendicular distance from the pivot Q to this horizontal line of action is simply the radius R. This creates a clockwise torque of magnitude F⋅R. If this applied force F is large enough, its torque will overcome the opposing counter-clockwise torque from gravity (mgR), and the wheel will successfully climb the step. Thus, the statement that it "never climbs" is false.
Option C
The Tricky Tangential Force at P
Option C is where the problem gets mathematically rigorous. It states a "constant force" is applied tangentially at P. In physics, a "constant force" typically implies a force vector that is constant in both magnitude and direction. Since it is applied tangentially at the leftmost point P initially, this force points vertically upwards.
As the wheel climbs and rotates by an angle θ, the material point P moves. However, because P is diametrically opposite to Q, the horizontal distance from Q to P is always twice the horizontal distance from Q to the center C.
The horizontal distance from Q to C is Rcosθ. Therefore, the horizontal distance from Q to P is 2Rcosθ. The torque of this constant vertical applied force is:
τapplied=F(2Rcosθ)
But we must also account for the torque due to gravity, which acts downwards from the center C:
τgravity=−mg(Rcosθ)
The net torque τ is the sum of these two:
τ=F(2Rcosθ)−mgRcosθ=(2F−mg)Rcosθ
Notice the crucial cosθ term. As the wheel climbs, the angle θ increases, which means cosθ decreases. Consequently, the net torque τ decreases continuously as the wheel climbs. Option (c) is absolutely correct!
Option D
The Normal Force at X
Finally, let's analyze a force applied normal to the circumference at the bottommost point X. This force points directly towards the center C.
Because the wheel is a rigid body, the angle between the radius QC and the radius XC is permanently fixed at 90∘. This geometric rigidity means that the perpendicular distance from the pivot Q to the line of action XC is always exactly R, regardless of how much the wheel has rotated!
So, the applied torque is constant: τapplied=F⋅R.
However, the net torque is:
τ=FR−mgRcosθ
As θ increases, cosθ decreases, making the negative gravity term smaller. Subtracting a smaller number means the overall net torque actually increases continuously. Therefore, option (d) is false.
Conclusion
By carefully distinguishing between the applied torque and the net torque, and rigorously tracking the lever arms as the wheel rotates, we confidently conclude that the correct options are (a) and (c).
