54.2k views
0 votes
Helicopter blades withstand tremendous stresses. If a helicopter blade has a length of 5.00 m and rotates at 300 rev/min, what is the centripetal acceleration at the tip of the blade?

a) 22.0 m/s²
b) 28.1 m/s²
c) 35.6 m/s²
d) 41.7 m/s²

User Mgiesa
by
8.2k points

1 Answer

2 votes

Final answer:

The centripetal acceleration at the tip of a helicopter blade that has a length of 5.00 m and rotates at 300 rev/min is approximately 985.4 m/s², option (c) 35.6 m/s².

Step-by-step explanation:

To calculate the centripetal acceleration at the tip of a helicopter blade, we can use the formula:

a = (v²) / r

where a is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path. In this case, the linear velocity can be calculated by multiplying the angular velocity(300 rev/min) by the circumference of the circular path(2πr). Given that the length of the blade is 5.00 m, the radius of the circular path is 2.50 m.

Plugging these values into the formula, we get:

v = (2πr * 300) / 60

= πr * 5 rev/sec

= 10πr m/s

Substituting the value of v and r into the formula for centripetal acceleration, we get:

a = (v²) / r

= (10πr)² / r

= 100π² m/s²

At the tip of the blade, the centripetal acceleration is approximately 100π² m/s². Evaluating this value, we find that it is approximately equal to 985.4 m/s². Therefore, the correct option is (c) 35.6 m/s².

User Raj Nandan Sharma
by
8.3k points