Question

a bus moving with a speed of 36 km per hour negotiates a level circular road of radius 10 M what should be the coefficient of friction so that the bus may not skid off

Answer 1

The question states level circular road of radius 10 i.e. accelaration in tangential direction which is now =2 m/s^2.

Since a body in circular path has two components of acceleration, the radial (towards centre) and tangential.

For the radial acceleration: A(r) = V^2÷R = 30^2÷500 = 900÷500 = 1.8 rad/sec^2.

These two Vectors have an angle of 90° b/w them, the net Vectorial acceleration becomes: (A(t)^2 +A(r)^2)^1÷2 (because cos 90°=0) =√7.24 ≈ 2.7 m/s^2

Answer 2

Answer:

Answer :

• $\twoheadrightarrow$ The question states level circular road of radius 10 i.e. accelaration in tangential direction which is now =2 m/s^2.

• $\twoheadrightarrow$ Since a body in circular path has two components of acceleration, the radial (towards centre) and tangential.

• $\twoheadrightarrow$ For the radial acceleration: A(r) = V^2÷R = 30^2÷500 = 900÷500 = 1.8 rad/sec^2.

• $\twoheadrightarrow$ These two Vectors have an angle of 90° b/w them, the net Vectorial acceleration becomes: (A(t)^2 +A(r)^2)^1÷2 (because cos 90°=0) =√7.24 ≈ 2.7 m/s^2