52.9k views
3 votes
You want to race a hoop, I_{hoop} = MR^2I hoop =MR 2 , a solid sphere, I_{solid sphere} = \frac{2}{5}MR^2I solidsphere = 5 2 MR 2 and a solid cylinder, I_{solid cylinder} = \frac{1}{2}MR^2I solidcylinder = 2 1 MR 2 down a ramp. Each object has the same mass and radius. Predict which object will win the race.A. The hoop will win.B. The solid sphere will win.C. The solid cylinder will win.D. All three objects will tie.E. It's impossible to tell

1 Answer

4 votes

Answer:

The fastest object is the sphere, so it is the winner

Step-by-step explanation:

To know which object will arrive faster down, let's look for the velocity of the center of mass of each object. Let's use the concept of mechanical energy

Highest point

Em₀ = U = mg y

Lowest point


Em_(f)= K =
K_(rot) +
K_(cm) = ½ I w² + ½ m
v_(cm)²

Angular velocity is related to linear velocity.

v = w r

w = v / r


Em_(f) = ½ I
v_(cm)²/r² + ½ m
v_(cm)²


Em_(f) = ½ (I / r² + m)
v_(cm)²

Energy is conserved

Em₀ =
Em_(f)

mg y = ½ (I / r² + m)
v_(cm)²


v_(cm) = √2 g y / (I / mr² +1)

With this expression we can know which object arrives as a higher speed, therefore invests less time and is the winner. Let's calculate the speed of the center of mass of each

Ring

I = m r²


v_(cm) = √ (2 g y / (m r² / mr² + 1))


v_(cm) = √ (2gy 1/2)


v_(cm) = (√ 2gy) 0.707

Solid sphere

I = 2/5 m r²


v_(cm) = √ (2gy / (2/5 m r² / mr² + 1)


v_(cm) = √ (2gy / (7/5))


v_(cm) = √ (2gy 5/7)


v_(cm) = (√ 2gy) 0.845

Cylinder

I = ½ m r²


v_(cm) = √ (2gy / ½ mr² / mr² + 1)


v_(cm) = √ (2gy / (3/2))


v_(cm) = √ (2g y 2/3)


v_(cm) = (√ 2gy) 0.816

The fastest object is the sphere, so it is the winner when descending the ramp

User Will Hancock
by
8.9k points

Related questions

asked May 26, 2024 55.5k views
Ameer Tamboli asked May 26, 2024
by Ameer Tamboli
8.9k points
1 answer
1 vote
55.5k views
asked Mar 21, 2024 961 views
Ahmed Shamel asked Mar 21, 2024
by Ahmed Shamel
8.4k points
1 answer
5 votes
961 views