Question #10e47

Question

Question #10e47

2 Answers

Impact of this question

2008 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-18T18:04:47

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

Explanation:

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

so, angular retardation= $α$ $alpha$ = change in angular velocity i.e $\frac{ω}{t}$ $omega/t$ = $\frac{40}{10}$ $40/10$ i.e $4$ $4$ radian/ $s^{2}$ $s^2$

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

moment of inertia of a cylinder passing through its axis is $\frac{M r^{2}}{2}$ $(Mr^2)/2$ = $1.5 K g . m^{2}$ $1.5 Kg.m^2$

so. $τ$ $tau$ = $1.5 \cdot 4 N . m$ $1.5*4 N.m$ i,e $6 N . m$ $6 N.m$

so,angular power required = $\frac{τ \cdot θ}{t}$ $(tau* theta)/t$ i.e about $120$ $120$ watt( $τ \cdot θ$ $tau*theta$ = work done,dividing by time required for this angular displacement of $θ$ $theta$ by the time $t$ $t$ in which it was brought to rest due to application of this torque) ( $θ$ $theta$ is angular displacement)( $θ = \frac{{(ω)}^{2}}{2 α} = \frac{40^{2}}{2 \cdot 4} = 200 r a d i a n$ $theta = (omega)^2/(2 alpha)=(40^2)/(2*4)=200 radian$ )

Answer 2 · 2018-01-18T18:07:14

The torque is $= 6 N m$ $=6Nm$ . The power is $= 240 W$ $=240W$

Explanation:

The torque and the angular acceleration are related

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

$τ = I \times α$ $tau=Ixxalpha$

The moment of inertia of the solid cylinder is

$I = \frac{m r^{2}}{2}$ $I=(mr^2)/2$

The mass of the cylinder is $m = 3 k g$ $m=3kg$

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

The moment of inertia is $I = \frac{3 \times 1^{2}}{2} = 1.5 k g m^{2}$ $I=(3xx1^2)/2=1.5kgm^2$

The initial angular velocity is $ω_{0} = 40 r a d s^{- 1}$ $omega_0=40rads^-1$

The final angular velocity is $ω_{1} = 0 r a d s^{- 1}$ $omega_1=0rads^-1$

The time is $t = 10 s$ $t=10s$

Apply the equation (to find the angular acceleration $(α)$ $(alpha)$ )

$ω_{1} = ω_{0} + α t$ $omega_1=omega_0+alphat$

$α = \frac{ω_{1} - ω_{0}}{t} = \frac{0 - 40}{10} = - 4 r a d s^{- 2}$ $alpha=(omega_1-omega_0)/t=(0-40)/10=-4rads^-2$

The torque to be applied is

$τ = I α = 1.5 \times 4 = 6 N m$ $tau=Ialpha=1.5xx4=6Nm$

$Power (W) = torque (Nm) \times angular velocity (rads^-1)$ $"Power (W)"= "torque (Nm)"xx "angular velocity (rads^-1)"$

$P = τ \times ω_{0}$ $P=tauxx omega_0$

The power is $P = 6 \times 40 = 240 W$ $P=6xx40=240W$

Science

Math

Humanities

... and beyond

Question #10e47

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question