Question

Question #10e47

Answer 1

Toque applied = `6 N.m` & power is about`120` watt

Explanation:

Here,by applying the torque angular retardation was done to stop its motion(considering no skidding was there).

so, angular retardation= `alpha` = change in angular velocity i.e `omega/t` = `40/10` i.e `4` radian/`s^2`

Now, torque =`Ialpha` (where, I is the moment of inertia)

moment of inertia of a cylinder passing through its axis is `(Mr^2)/2` =`1.5 Kg.m^2`

so. `tau` =`1.5*4 N.m` i,e `6 N.m`

so,angular power required = `(tau* theta)/t` i.e about `120` watt(`tau*theta`= work done,dividing by time required for this angular displacement of `theta` by the time `t` in which it was brought to rest due to application of this torque) ( `theta` is angular displacement)(`theta = (omega)^2/(2 alpha)=(40^2)/(2*4)=200 radian`)

Answer 2

The torque is `=6Nm`. The power is `=240W`

Explanation:

The torque and the angular acceleration are related

`"Torque (Nm)"="Moment of inertia (kgm^2)"xx "angular acceleration (rads^-2)"`

`tau=Ixxalpha`

The moment of inertia of the solid cylinder is

`I=(mr^2)/2`

The mass of the cylinder is `m=3kg`

The radius of the cylinder is `r=1.0m`

The moment of inertia is `I=(3xx1^2)/2=1.5kgm^2`

The initial angular velocity is `omega_0=40rads^-1`

The final angular velocity is `omega_1=0rads^-1`

The time is `t=10s`

Apply the equation (to find the angular acceleration `(alpha)`)

`omega_1=omega_0+alphat`

`alpha=(omega_1-omega_0)/t=(0-40)/10=-4rads^-2`

The torque to be applied is

`tau=Ialpha=1.5xx4=6Nm`

`"Power (W)"= "torque (Nm)"xx "angular velocity (rads^-1)"`

`P=tauxx omega_0`

The power is `P=6xx40=240W`