Question

For the angular velocity vs. time graph depicted above if both objects roll without slipping and the displacement of both objects is the same, at time, $t$, which of the following statements with regard to the radius of the red and blue objects is correct?

• A. The radius of the blue object is 3 times larger than the blue object
• B. The radius of the blue object is $2/3$ the size of the red object
• C. Both objects have the same radius since the displacements are the same
• D. The radius of the blue object is $\frac{8}{5}$ the radius of the red object
• E. All of the statements are false. The graph indicates that the angular displacements are not the same, there for the displacements cannot be the same

Answer 1

Answer: (B)

The radius of the blue object is $2/3$ the size of the red object

Area under the w-t graph gives the angular displacement of the object during that period.

For Blue object : $AB=PQ=w$ $OB=\frac{t}{2}$ $BQ=\frac{t}{2}$

$\therefore$ Angular displacement of blue object in time t, ${\theta }_b=$ Area of triangle(OAB) + Area of rectangle(APQB)

$\therefore$ ${\theta }_b=\frac{1}{2}\left(w\right)\times \frac{t}{2}+w\times \frac{t}{2}=\frac{3wt}{4}$

For red object : $OL=\frac{w}{2}$ $OQ=t$

$\therefore$ Angular displacement of red object in time t, ${\theta }_r=$ Area of rectangle(LMQO)$=\frac{wt}{2}$

As linear displacement $S=R\theta$ where $R$ is the radius of the object

Given: $S_r=S_b$ i.e $R_r{\theta }_r=R_b{\theta }_b$

$\therefore$ $R_r\times \frac{wt}{2}=R_b\times \frac{3wt}{4}$ $⟹R_b=\frac{2}{3}{R}_r$