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Dynamics of Circular Motion

A turn of radius 20 {~m} is banked for the vehicle of mass 200

Question

Medium

Solving time: 3 mins

A turn of radius $20 m$ is banked for the vehicle of mass $200 kg$ going at a speed of $10 m / s$ . Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed
(a) $5 m / s$
(b) $15 m / s$ .
Assume that friction is sufficient to prevent slipping. $(g = 10 m / s^{2})$

Name: Video solution 1: A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed(a) 5 m/s(b) 15 m/s.Assume that friction is sufficient to prevent slipping. (g=10 m/s2)
Uploaded: 2022-09-28T06:36:09.000Z
Duration: 7 min
Description: Video solution 1: A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed(a) 5 m/s(b) 15 m/s.Assume that friction is sufficient to prevent slipping. (g=10 m/s2)

Views: 5,742 students

Updated on: Sep 28, 2022

Book:

Mechanics Volume 1 (DC Pandey)

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Liam Discussed

A turn of radius

20 m

is banked for the vehicle of mass

200 kg

going at a speed of

10 m / s

. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed
(a)

5 m / s

(b)

15 m / s

.
Assume that friction is sufficient to prevent slipping.

(g = 10 m / s^{2})

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

(a)

The turn is banked for speed

v = 10 m / s

Therefore,

tan θ = \frac{v ^{2}}{r g} = \frac{( 10 ) ^{2}}{( 20 ) ( 10 )} = \frac{1}{2}

Now, as the speed is decreased, force of friction

f

acts upwards.
Using the equations

Σ F_{x} = \frac{m v ^{2}}{r}

and

Σ F_{y} = 0

, we get

N sin θ - f cos θ = \frac{m v ^{2}}{r}

N cos θ + f sin θ = m g

Substituting,

θ = tan^{- 1} (\frac{1}{2}), v = 5 m / s, m = 200 kg

and

r = 20 m

, in the above equations, we get

f = 3005 N

(upwards)
(b) Solution Image

In the second case force of friction

f

will act downwards.
Using

Σ F_{x} = \frac{m v ^{2}}{r}

and

Σ F_{y} = 0,

we get

N sin θ + f cos θ = \frac{m v ^{2}}{r}

N cos θ - f sin θ = m g

Substituting

θ = tan^{- 1} (\frac{1}{2}), v = 15 m / s, m = 200 kg

and

r = 20 m

in the above equations, we get

f = 5005 N

(downwards)

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Question Text	A turn of radius $20 m$ is banked for the vehicle of mass $200 kg$ going at a speed of $10 m / s$ . Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed (a) $5 m / s$ (b) $15 m / s$ . Assume that friction is sufficient to prevent slipping. $(g = 10 m / s^{2})$
Updated On	Sep 28, 2022
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	281
Avg. Video Duration	14 min

Text solutionVerified

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A turn of radius 20 m is banked for the vehicle of mass 200 kg going at a speed of 10 m/s. Find the direction and magnitude of frictional force acting on a vehicle if it moves with a speed(a) 5 m/s(b) 15 m/s.Assume that friction is sufficient to prevent slipping. (g=10 m/s2)

Text solutionVerified

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