Question

A solid sphere of mass 0.1 kilogram and radius 2 cm rolls down on an inclined plane 1.4 meter in length of slope 1 in 10. Starting from rest its velocity (m/s) will be?

Answer 1

equating (1) and (4) we get

m

g

h

=

1

2

m

v

2

+

1

2

I

ω

2

......(5)

Using (3) and (4), equation (5) becomes

m

g

h

=

1

2

m

(

r

ω

)

2

+

1

2

×

1

2

m

r

2

ω

2

⇒

g

h

=

(

1

2

+

1

4

)

r

2

ω

2

⇒

g

h

=

3

4

r

2

ω

2

Solving for

r

⇒

r

=

√

4

g

h

3

ω

2

Inserting given values we get value of radius

r

as

r

=









⎷

4

×

10

×

3

3

(

2

√

2

)

2

r

=

√

5

m

Answer 2

Answer:

The velocity will be 1.4 m/s.

Explanation:

Given that,

Mass of solid sphere = 0.1 kg

Radius = 2 cm

Inclined plane = 1.4 m

Length of slop is given by

$\tan\theta=\dfrac{1}{10}$

$\theta=\tan^{-1}(\dfrac{1}{10})$

$\theta=0.099$

So, $\sin\theta=0.099$

The height of the incline is

$h'=h\sin\theta$

Put the value into the formula

$h'=0.099\times`1.4$

$h'=0.14\ m$

We need to calculate the velocity

Using conservation of energy

$mgh'=\dfrac{1}{2}mv^2+\dfrac{1}{5}mv^2$

Put the value into the formula

$9.8\times0.14=\dfrac{7}{10}v^2$

$v^2=\dfrac{10\times9.8\times0.14}{7}$

$v^2=1.96$

$v=\sqrt{1.96}$

$v=1.4\ m/s$

Hence, The velocity will be 1.4 m/s.