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Q27 PE

College Physics (Urone)
Found in: Page 224

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Short Answer

(a) What is the radius of a bobsled turn banked at 75.0°and taken at 30.0 m/s, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you?

(a) The radius is 24.6m.

(b) 36.6 m/s is centripetal acceleration.

(c) Yes, the centripetal acceleration is 3.7 times greater than gravity's acceleration.

See the step by step solution

Step by Step Solution

Step 1: Definition of Centripetal acceleration

The rate of change of transverse velocity is known as centripetal acceleration.

Step 2: Calculating radius of bobsled turn

(a) The ideal velocity for a banked road is,


Here, R is the radius of the turn, g is the acceleration due to gravity, and θ is the ideal angle of banking.

Rearranging the above equation in order to get an expression for the radius of the turn,


Substitute 30.0 m/s for v,9.8 m/s2 for g, and 75.00 for θ,


Here, the radius of the bobsled turn is =24.6m

Step 3: Calculating centripetal acceleration

(a) The centripetal acceleration is,

ac = v2r

Substitute 30.0 m/s for v , and 24.6 m for r,

ac = 30 m/s224.6 m = 36.6 m/s2

Hence, the centripetal acceleration is 36.6 m/s.

Step 4: Determining is acceleration seem larger

The proportion of centripetal to gravitational acceleration is,

acg = 36.6 m/s29.8 m/s2 = 3.7 = 3.7g

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