Answer to Question #174287 in Mechanics | Relativity for Jay

Question

Find the linear velocities and accelerations of centers of sphere, disc and hoop that roll down an inclined plane without slipping. The incline of height h =1m makes an angle of 300 to the horizontal. The initial velocity of all objects v0 =0, compare calculated velocities and accelerations with the velocity and acceleration of the box, which slides from this incline without friction.

Answer 1

Explanations & Calculations

• By applying F=ma along the moving direction & using $\small \tau=I\alpha$ simultaneously (as the rolling object does not slip, friction helps it roll by the generated torque), linear acceleration can be found & then using $\small V^2=U^2+2as$, linear velocity can be found as the linear acceleration is already found & it is constant for the given object.
• The length of the incline is

$\qquad\qquad \begin{aligned} \small s&=\small \frac{1}{\sin30}=\bold{2m} \end{aligned}$

• For the sphere (a hollow sphere)

$\qquad\qquad \begin{aligned} \small I_s&=\small \frac{2mr^2}{3}\\ \small mg\sin\theta-f&=\small ma\cdots(1)\\ \small f\cdot r&=\small \frac{2mr^2}{3}\cdot\frac{a}{r}\\ \small f&=\small \frac{2ma}{3}\cdots(2) \\ \small\text{By (1) and (2)}\\ \therefore \small a_s&=\small \frac{3}{5}\cdot g\sin\theta=\bold{2.94ms^{-2}}\\ \\ \small V_s&=\small \sqrt{\frac{6g}{5}}=\bold{3.429ms^{-1}} \end{aligned}$

• For the disc

$\qquad\qquad \begin{aligned} \small I_d&=\small \frac{mr^2}{2}\\ \small mg\sin\theta-f&=\small ma\cdots(1)\\ \small f\cdot r&=\small \frac{mr^2}{2}\cdot\frac{a}{r}\\ \small f&=\small \frac{ma}{2}\cdots(2) \\ \small\text{By (1) and (2)}\\ \therefore \small a_d&=\small \frac{2}{3}\cdot g\sin\theta=\bold{3.26ms^{-2}}\\ \\ \small V_d &=\small \sqrt{\frac{4g}{3}}=\bold{3.615ms^{-1}} \end{aligned}$

• For the hoop

$\qquad\qquad \begin{aligned} \small I_h&=\small mr^2\\ \\ \small mg\sin\theta-f&=\small ma\cdots(1)\\ \small f\cdot r&=\small mr^2\cdot\frac{a}{r}\\ \small f&=\small ma\cdots(2) \\ \small\text{By (1) and (2)}\\ \therefore \small a_h&=\small \frac{1}{2}\cdot g\sin\theta=\bold{2.45ms^{-2}}\\ \\ \small \text{Then,}\\ \small V_h&=\small \sqrt{g}=\bold{3.13ms^{-1}} \end{aligned}$

• For the block

$\qquad\qquad \begin{aligned} \small mg\sin\theta&=\small ma\\ \small a_b&=\small g\sin\theta=\bold{4.9ms^{-2}}\\ \\ \small \text{then,} \\ \small V_b&=\small \sqrt{4g\sin\theta}=\bold{4.43ms^{-1}} \end{aligned}$

• Comparison

$\qquad\qquad \begin{aligned} \small I_b<I_d&<I_s<I_h\\ \small a_b>a_d&>a_s>a_h\\ \small V_b>V_d&>V_s>V_h \end{aligned}$