Question

Consider a pure-rolling race where two objects A and B are simultaneously released from the top of an inclined plane and roll down to a finish line. For each group below, answer which object will win the race or it will be a tie (A or B or Tie).
1.A: 100g solid cylinder with R = 10cm vs B: 200g solid cylinder with R = 10cm.
2. A: 100g solid cylinder with R = 10cm vs B: 100g solid cylinder with R = 20cm.
3. A: 100g solid cylinder with R = 10cm vs B: 100g hollow cylinder with R = 10cm.
4. A: 100g solid cylinder with R = 10cm vs B: 100g solid sphere with R = 10cm.

Answer 1

Explanation:

Given that,

(1). Mass of solid cylinder A= 100 g

Radius = 10 cm

Mass of solid cylinder B= 200 g

Radius = 10 cm

(2). Mass of solid cylinder A= 100 g

Radius = 10 cm

Mass of solid cylinder B= 100 g

Radius = 20 cm

We need to calculate the moment of inertia of solid cylinder A and B

Using formula of moment of inertia

$I=\dfrac{1}{2}mR^2$

But we know that,

For pure rolling rotation,

An object under the influence of gravity does not depend on the object's mass and its size.

Therefore, Both solid cylinders reach on same time.

(3). Mass of solid cylinder A= 100 g

Radius = 10 cm

Mass of hollow cylinder B= 100 g

Radius = 10 cm

We need to calculate the moment of inertia of solid cylinder A

Using formula of moment of inertia

$I=\dfrac{1}{2}mR^2$

$I_{A}=\dfrac{1}{2}\times100\times10^{-3}\times(10\times10^{-2})^2$

$I_{A}=0.0005\ kg-m^2$

We need to calculate the moment of inertia of hollow cylinder B

Using formula of moment of inertia

$I=mR^2$

$I_{B}=100\times10^{-3}\times(10\times10^{-2})^2$

$I_{B}=0.001\ kg-m^2$

Here, The moment of inertia of A is less for hollow cylinder.

Therefore, Solid cylinder will reach first.

(4). Mass of solid cylinder A= 100 g

Radius = 10 cm

Mass of solid sphere B= 100 g

Radius = 10 cm

We need to calculate the moment of inertia of solid cylinder A

Using formula of moment of inertia

$I=\dfrac{1}{2}mR^2$

$I_{A}=\dfrac{1}{2}\times100\times10^{-3}\times(10\times10^{-2})^2$

$I_{A}=0.0005\ kg-m^2$

We need to calculate the moment of inertia of solid sphere B

Using formula of moment of inertia

$I=\dfrac{2}{5}mR^2$

$I_{B}=\dfrac{2}{5}\times100\times10^{-3}\times(10\times10^{-2})^2$

$I_{B}=0.0004\ kg-m^2$

Here, The moment of inertia of A is more for solid sphere.

Therefore, Solid sphere will reach first

Hence, This is the required solution.