Question

10. A solid cylinder of diameter 30 cm at the top of an
2 inclined plane 2.0 m high is released and rolls down
to the inclined plane without loss ofenergy due to friction.
To Its linear speed at the bottom is
(a) 5.29m/sec b) 4.10x 10ºm/sec
51 m/sec d) 51 cm/sec​

Answer 1

Dear Student,

◆ Answer -

(a) 5.29m/sec

● Explanation -

When the solid cylinder rolls down, its potential energy is converted to its kinetic energy (rotational & translational).

mgh = 1/2 mv^2 + 1/2 Iw^2

gh = 1/2 (rw)^2 + 1/2 (1/2 r^2)w^2

gh = 3/4 (rw)^2

gh = 3/4 v^2

v = √(4gh/3)

Substitute values,

v = √(4 × 9.8 × 2 / 3)

v = 5.11 m/s

Therefore, linear speed is 5.11 m/s.

Thanks dear...

Answer 2

Answer:

The linear speed at the bottom is 51 cm/s.

(d) is correct option.

Explanation:

Given that,

Diameter = 30 cm

Height = 2.0 m

We need to calculate the total kinetic energy at bottom

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$....(I)

The moment of inertia is

$I= \dfrac{mr^2}{2}$

Put the value in the equation (I)

$K.E=\dfrac{1}{2}\times mv^2+\dfrac{1}{2}\times\dfrac{mr^2}{2}\times(\dfrac{v}{r})^2$

$K.E=\dfrac{3}{4}mv^2$

Potential energy at top = mgh

Therefore,

$mgh=K.E$

Put the value into the formula

$mgh=\dfrac{3}{4}mv^2$

$9.8\times2.0=\dfrac{3}{4}v^2$

$v^2=\dfrac{9.8\times2.0\times4}{3}$

$v=\sqrt{\dfrac{9.8\times2.0\times4}{3}}$

$v=5.1\ m/s$

$v= 51\ cm/s$

Hence, The linear speed at the bottom is 51 cm/s.