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a hoop of mass 1.2 kg and radius 65 cm has a string wrapped around its circumference and lies flat on a horizontal frictionless table. the string is pulled with a constant force 7 n. how far does the center of the hoop travel in 4 s? answer in units of m. what is the angular velocity of the hoop about its center of mass after 4

Sagot :

1.57 rad/s is the angular velocity of the hoop about its center of mass.

Solution:

Given that

Weight of hoop m = 1.2 kg

radius of  hoop R = 65 cm = 0.65 m

Force = 7 N

Torque on the hoop Τ = F×R        →equation (1)

Where F is applied force on hoop

We also know that

Τ = Iα[tex]R^{2}[/tex]

Where α is the angular acceleration

I is the moment of inertia of hoop

For hoop    I = m[tex]R^{2}[/tex]

T =  (m[tex]R^{2}[/tex])α           →equation (2)

From equation 1 and 2

FR = (m)α

F = mRα

α = F/mR

α = 7/1.2×0.65 = 3.79 rad/s×s

α= 3.79 rad/s×s

We have to calculate distance covered  by center of hoop in 5 s

d = (1/2)at×t                 →equation(3)

Relation between linear acceleration and radius is

a = αR

a = 3.79×0.65

a = 2.46 m/s×s

now from equation (3)

d = 0.5×2.46×4×4

d = 19.68 m

so the  distance covered by center of hoop in 4 s is 19.68 m.

Part(2)

angular velocity of hoop

angular momentum about the center of the hoop

w = v/R

ωt = 2×π

t is  given 4 s

ωt   = 6.28

ω×4= 6.28

ω = 1.57 rad/s

Hence,  1.57 rad/s is the angular velocity of the hoop about its center of mass.

Learn more about angular velocity here:- https://brainly.com/question/6860269

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