Question

A small particle initially at point p starts moving from rest.

Answer 1

Answer:

Initial height is r(1−cosα)

Final height is r(1−sinβ)

Net vertical displacement is r(1−sinβ)−r(1−cosα)=r(cosα−sinβ)

Apply equation of kinematic

v

2

−u

2

=2as

v

2

=2gr(cosα−sinβ)

Normal reaction,

N=mgsinβ−

r

mv

2

0=mgsinβ−

r

m2gr(cosα−sinβ)

3sinβ=2cosα

Hence, the relation is 3sinβ=2cosα

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Answer 2

The total time is equal to 13 seconds

Given:

A small particle initially at point p starts moving from rest. It moves over the first part as 12m/s2 and remaining retardation is 3ms2

To Find:

When the particle starts from point p and stops at point Q .To get the time taken to complete the whole journey

Solution:

From the equation of motion,

We  get,

$v^2 = u^2+2as$

v^2 =  0 + 2(12)(15)

$v0^2 =$ 360

For the first half we get,

From the equation of motion,

We write the first equation as,

s = $ut+1/2at^2$

15 =  0 + $1/2(12)(t1^2)$

$t1^2 =$ 2.5

t1 = 1.58sec

For the second half,

s = $ut+1/2at^2$

15 = v0t2 - $1/2 (3)(t2^2)$

15 = 19 ( t2)  -1.5 $(t2^2)$

1.5 $(t2^2)$ -19t2+15 =0

Total time taken,

t1+t2 = 11.82 + 1.58sec

= 13seconds

Hence, the total time taken is 13 seconds

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