Question 750: Centre of Mass and Rotational Motion

Question

Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figure) starting at A and C with linear speeds v
1
and v
2
, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and v
1
= 3 m/s, then v
2
in m/s is (g = 10 m/s
2
)

Accepted Answer

Suppose mass and radius of each disc are m and R respectively. Also potential energy at points B and D is zero i.e., they are on reference line.
Given final kinetic energy for each disc is same, say it is K.
Applying energy conservation principle,
For surface AB,
$${1 \over 2}{I_2}\omega _1^2 + mg 	imes 30 = K$$ ..... (i)
For surface CD,
$${1 \over 2}{I_2}\omega _2^2 + mg 	imes 27 = K$$ ..... (ii)
From eqns. (i) and (ii), we get
$${1 \over 2}{I_2}\omega _1^2 + mg 	imes 30 = {1 \over 2}{I_2}\omega _2^2 + mg 	imes 27$$ ...... (iii)
Here, $${\omega _1} = {{{v_1}} \over R}$$, $${\omega _2} = {{{v_2}} \over R}$$, v
1
= 3 m s
$$-$$1
, v
2
= ?
I
1
= I
2
= Moment of inertia of disc about the point of contact $$ = {1 \over 2}m{R^2} + m{R^2} = {3 \over 2}m{R^2}$$
From eqn. (iii), $${1 \over 2}\left( {{3 \over 2}m{R^2}} ight) 	imes {\left( {{3 \over R}} ight)^2} + m 	imes 10 	imes 30$$
$$ = {1 \over 2}\left( {{3 \over 2}m{R^2}} ight) 	imes {\left( {{{{v_2}} \over R}} ight)^2} + m 	imes 10 	imes 27$$
$${{27} \over 4} + 300 = {3 \over 4}v_2^2 + 270$$
$${3 \over 4}v_2^2 = {{27} \over 4} + 30 \Rightarrow 3v_2^2 = 147$$
$$v_2^2 = 49$$ $$	herefore$$ v
2
= 7 m s
$$-$$1

Centre of Mass and Rotational Motion

Integer / Numeric Answer Type