Question

6. A device for acclimating military pilots to the high accelerations they must experience consists of...

6. A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 35.1 m/s2 with a beam of length 5.39 m, what rotation frequency is required?

? Hz

7. You want to design an oval racetrack such that 3200 lb racecars can round the turns of radius 1000 ft at 105 mi/h without the aid of friction. You estimate that when elements like downforce and grip in the tires are considered the cars will round the turns at a maximum of 175 mi/h. Find the banking angle θ necessary for the racecars to navigate these turns at 105 mi/h and without the aid of friction.

? degree

This banking and radius are very close to the actual turn data at Daytona International Speedway where 3200 lb stock cars travel around the turns at about 175 mi/h. What additional radial force is necessary to hold the racecar on the track at 175 mi/h?

? N

8. A 5.21-kg ball hangs from the top of a vertical pole by a 2.51-m-long string. The ball is struck, causing it to revolve around the pole at a speed of 4.83 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 m/s2.

? degree

What is the tension of the string?

? N

Homework Answers

Answer #1

6)

given

a_rad = 35.1 m/s^2

L = r = 5.39 m

we know,

a_rad = r*w^2

==> w = sqrt(a_rad/r)

= sqrt(35.1/5.39)

= 2.55 hz

8)

let T is the tension i the string and theta is the angle made by the string with vertical.

Fnety = 0

T*cos(theta) - m*g = 0

T*cos(theta) = m*g -----------(1)

Fnetx = m*a_rad

T*sin(theta) = m*v^2/r

T*sin(theta) = m*4.83^2/(L*sin(theta)

T*sin^2(theta) = m*4.83^2/2.51 -----(2)

take equation(2)/equation(1)

sin^2(theta)/cos(theta) = 4.83^2/(2.51*g)

(1 - cos^2(theta))/cos(theta) = 4.83^2/(2.51*9.8)

let cos(theta) = x

(1 - x^2)/x = 4.83^2/(2.51*9.8)

x = 0.632

cos(theta) = 0.632

theta = cos^-1(0.632)

= 51 degrees <<<<<<<<<<<-------------Answer


from equation 1

T = m*g/cos(theta)

= 5.21*9.81/(cos(51))

= 81.2 N <<<<<<<<<<<-------------Answer

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