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

Short Answer

Expert verified

(a) The radius is 24.6m.

(b) 36.6 m/sis centripetal acceleration.

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

Step by step solution

01

Definition of Centripetal acceleration

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

02

Calculating radius of bobsled turn

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

v=Rgtanθ

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

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

R=v2gtanθ

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

R=(30.0m/s)2(9.8m/s2)×tan(75.00)=24.6m

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

03

Calculating centripetal acceleration

(a) The centripetal acceleration is,

ac=v2r

Substitute 30.0 m/sfor v , and 24.6 mfor r,

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

Hence, the centripetal acceleration is 36.6 m/s.

04

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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