Question

A rope, to which a weight is attached, passes around a pulley 50 cm in diameter. The angular acceleration of the pulley is 18 rad/s2. If the pulley is initially rest, find
a- The time required for the weight to attain a velocity of 15 m/s
b- The number of revolution through which the pulley rotates during that period,
c- The total acceleration of a point on the rim of the pulley 0.5 second after it was at rest.

Answer 1

Given:

A rope, to which a weight is attached, passes around a pulley 50 cm in diameter. The angular acceleration of the pulley is 18 rad/s2. The pulley is initially at rest.

To find:

• Time required for weight to achieve 15 m/s velocity.
• Number of revolutions during that period
• Total acceleration of a point on the rim of pulley after 0.5 sec after starting.

Calculation:

Linear Acceleration of the pulley :

$a = \alpha \times r$

=> a = 18 × 0.25

=> a = 4.5 m/s².

So , let time be t :

Applying 1st Equation of Kinematics:

v = u + at

=> 15 = 0 + {(4.5) × t}

=> t = 15/(4.5)

=> t = 3.33 seconds.

So time taken to achieve the required velocity is 15 seconds.

Now , total number of rotations be R

R = (displacement)/(circumference)

=> R = (½$\alpha$t² × r)/(2πr)

=> R ={ ½ × 18 × (3.33)²} /2π

=> R = 15.88

=> R $\approx$ 16 rotations

So, number of rotations will be 16.

For pulley , linear acceleration is 4.5 m/s²

Linear velocity be v = 0 + (4.5×½)

=> v = 2.25 m/s

Hence centripetal acceleration will be :

$a_{c}$ = v²/r = (2.25)²/0.25 = 20.25 m/s²

So net acceleration will be vector addition of linear acceleration and centripetal acceleration:

acc = √{(4.5)² + (20.25)²}

=> acc. = √(430.31)

=> acc. = 20.74 m/s².

So net acceleration is 20.74 m/s²

Answer 2

Answer:

r=.25 m

@=18 rad/s2

Wo=0

Vo=0

sol (a)

t=??

V=15 m/s

V=Vo+ a*t  =====>1

15=0+a*t

a=r*@

a=25*18=4.5

=> in 1

15==4.5*t

==> t=3.33 s

________________________

sol(b)

n=??

n=2*TI *W

W=Wo+@t

  =0+18*3.33

  =60 rad/s

==> n=2*TI*60=377 rev/s

_____________________________

sol(c)

acc=??

a=جذر{ a1^2+a2^2}

V=Vo+a*t

V=0+4.5*0.5

V=2.25 m/s

a=v^2/r

a=2.25^2/25

a=20.25 m/s^2

==> acc={ 4.5^2 +20.25^2}=20.74 m/s^2

Explanation: