WPE O11 - Physics.lt

Solution 11

Solution to Question 11

1. Analyzing the Forces and Work

The disk moves on a frictionless horizontal floor. The only horizontal force acting on the disk is the tension $T$ from the cord. Since the cord is inextensible and always tangent to the cylinder as it winds, the tension force is always perpendicular to the velocity vector $\vec{v}$ of the disk (which is tangential to the path).

$$ P = \vec{F} \cdot \vec{v} = \vec{T} \cdot \vec{v} = T v \cos(90^\circ) = 0 $$

Since the power delivered by the tension is zero, there is no work done on the disk. According to the Work-Energy Theorem, the change in kinetic energy is zero.

$$ \frac{1}{2}mv^2 = \text{constant} \implies v = \text{constant} $$

Thus, the speed remains constant.

2. Analyzing the Tension

The tension $T$ provides the centripetal force required for the circular motion (approximated as instantaneous circular motion with radius $r$).

$$ T = \frac{mv^2}{r} $$

As the cord winds around the cylinder, the effective length of the cord (the radius of curvature $r$) decreases ($r$ decreases). Since $m$ and $v$ are constant, and $r$ is in the denominator:

$$ r \downarrow \implies T \uparrow $$

Thus, the tensile force will increase.

Conclusion

The speed remains constant, and the tensile force increases.

Correct Option: (d)