Physics Problems & Solutions: Classical Mechanics

A particle is initially at rest at the top of a curved frictionless track. The x- and y-coordinate of the track are related in dimensionless units by y = x²/4, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what its tangential acceleration?

A. 0
B. g
C. gx/2
D. gx/√(x²+4)
E. (gx²)/√(x²+16)

(GR8677 #06)

Solution:

A. FALSE.
The particle slides down the curved track → the tangential acceleration is not zero.

B. FALSE.
The particle slides down the curved track, not in the vertical downward direction (the direction of gravity acceleration) → the tangential acceleration is not equal to g.

C. FALSE.
The unit of gx/2 is not the unit of acceleration.

D. TRUE.
The unit of gx/√(x²+4) = g → the unit of acceleration.

E. FALSE.
The unit of (gx²)/√(x²+16) is not the unit of acceleration.

Answer: D

Calculation:

$\\a_\text{tan}=g \sin \alpha\\\\\tan \alpha = \frac{dy}{dx}=\frac{x}{2}$

$\\\frac{\sin \alpha}{\cos \alpha}=\frac{\sin \alpha}{\sqrt{1-\sin^2\alpha}}=\frac{x}{2}\\\\\rightarrow \frac{\sin^2 \alpha}{1-\sin^2\alpha}=\frac{x^2}{4}\\\\\rightarrow4\sin^2 \alpha=x^2-x^2\sin^2 \alpha\\\rightarrow\sin^2 \alpha=\frac{x^2}{4+x^2}\\\rightarrow \sin \alpha=\frac{x}{\sqrt{4+x^2}}$

$\rightarrow a_{\tan}=\frac{gx}{\sqrt{4+x^2}}$

Physics Problems & Solutions

Classical Mechanics - Kinematics

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