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3 years ago

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A string is wrapped around a disk of mass 1.7 kg and radius 0.11 m. Starting from rest, you pull the string with a constant forc

e 7 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.15 m, your hand has moved a distance of 0.22 m.
a. At this instant, what is the speed of the center of mass of the disk?
b. At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?

2 answers:

V125BC [204]3 years ago

5 0

Answer:

Explanation:

a) work done in moving the disk by 0.15 m = its kinetic energy = 0.5 mv²

Fd = 0.5 mv²

2 fd = mv²

√( $\frac{2fd}{m}$ ) = v

v = 1.11 m/s

b) rotational kinetic energy = F ( 0.22 m - 0.15 m) = 0.49 J

Evgen [1.6K]3 years ago

5 0

Answer:

(a) Vcm = 1.43m/s

(b) KEcm = 2.59J

Explanation:

Given

m = mass of the solid disk = 1.7kg

Radius R = 0.11m

F = 7N

Let the distance moved by the the center of mass be x = 0.15m

And the distance moved by in unwinding the rope be d = 0.22m

The total workdone on the disk is causing it to rotate and also move its center of mass is F×(d+x)

W = 7×(0.15+0.22)

= 7× 0.37 = 2.59J

By work-energy theorem,

W = ΔKE = 1/2mVcm² + 1/2×I×ω²....(1)

I = moment of inertia = 1/2MR² and

ω = angular speed = Vcm/R

So substituting these expressions into the equation above we have

W = 1/2MVcm² + 1/2×1/2MR²×(Vcm/R)²

W = 1/2MVcm² + 1/4MVcm²

W = 3/4MVcm²

Vcm² = 4/3×W/M

Vcm = √(4/3×W/M)

Vcm = √(4/3×2.59/1.7)

Vcm = 1.43m/s

KE = workdone = 2.59J.

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Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

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3 years ago