Ask question

Ask question

All categories

1 month ago

12

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

edge of the disk. A friction pad exerts a force of 9.7 N on the outside of the disk. A cyclist is pedaling, spinning the disk at a typical 180 rpm. If she stops pedaling, how long will it take for the flywheel to come to a stop?

1 answer:

Keith_Richards [3.2K]1 month ago

7 0

Response:

Clarification:

Provided

weight of disk $m=12.5 kg$

diameter of disc $R=0.23 m$

weight of ring $m_r=7 kg$

Force $F=9.7 N$

$N=180 rpm$

$\omega =\frac{2\pi N}{60}$

$\omega =6\pi rad/s$

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

$=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2$

$=13.25\times 0.23^2=0.7009 kg-m^2$

At this point, Torque is $T=F\times R=I\cdot \alpha$

$9.7\times 0.23=0.7\times \alpha$

$\alpha =3.18 rad/s^2$

Utilizing $\omega _f=\omega +\alpha t$

$\omega _f=0$ in this scenario

$0=6\pi -3.18\times t$

$t=\frac{6\pi }{3.18}$

$t=5.92 s$

You might be interested in

A child runs at speed vc while an adult runs at speed va. if the adult gives the child a head start of distance d, how long will

ValentinkaMS [3465]

The child runs at speed vc.
The adult runs at speed va.
<span>The adult allows the child a lead of distance d.
Speed is calculated by distance divided by time: speed = distance/time.
To find the time it takes for the adult to catch the child, use time = d/v.
</span>Let tc be the child's time and ta the adult's time.
tc = D/vc, with D = D + d, so D + d = tc × vc, therefore D = tc × vc - d.
ta = D/va, where D = ta × va.
Equating both: tc × vc - d = ta × va
Thus, ta = (tc × vc - d) / va

4 0

2 months ago

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w

serg [3582]

Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

E = $\frac{\sigma}{2 \epsilon_{o}}$

where, $\sigma$ is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

F = eE

where, e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

F = ma

a = $\frac{eE}{m}$

= $\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}$

= $\frac{e \sigma}{2m \epsilon_{o}}$

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

$v_{f} = v_{i} + at$

= $(0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t$

= $\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}$

= 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

v' = $\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}$

= $\sqrt{921600 + 467820.43}$

= $\sqrt{1389420.43}$

= 1178.73 m/s

Consequently, we conclude that the proton's speed is 1178.73 m/s.

3 0

1 month ago

A system expands from a volume of 1.00 l to 2.00 l against a constant external pressure of 1.00 atm. what is the work (w) done b

Keith_Richards [3271]

The amount of work performed by a system at consistent pressure is defined by the following equation:
$W=p \Delta V = p (V_f - V_i)$
where
p represents pressure
$V_f$ as the final volume
$V_i$ as the initial volume

Plugging the values given in this case into the formula gives us
$W=p (V_f -V_i)=(1.00 atm)(2.00 L-1.00 L)=1.00 L\cdot atm$

Considering that $1 atm \cdot L = 101.3 J$ , the result for the work done becomes
$W= 1.00 atm \cdot L = 101.3 J$

8 0

1 month ago

Read 2 more answers

The surface is tilted to an angle of 37 degrees from the horizontal, as shown above in Figure 3. The blocks are each given a pus

inna [3103]

Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

The velocity of the center of mass for the two-block system just prior to their collision is 2.9489 m/s

Explanation:

Provided information:

θ = angle of the surface = 37°

m = mass of each block = 0.2 kg

v = speed = 0.35 m/s

t = collision time = 0.5 s

Question: What is the velocity of the center of mass for the two-block system right before the blocks collide, vf =?

Change in momentum:

$delta(P)=F*delta(t)$

$P_{f} -P_{i}=F*delta(t)$

$2m(v_{f} -v_{i})=F*delta(t)$

$v_{i} =0.35-0.35=0$

It’s essential to calculate the required force:

$F=(m+m)*g*sin\theta$

Here, g = acceleration due to gravity = 9.8 m/s²

$F=(0.2+0.2)*9.8*sin37=2.3591N$

$v_{f} =\frac{F*delta(t)}{2m} =\frac{2.3591*0.5}{2*0.2} =2.9489m/s$

6 0

2 months ago

A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .

Maru [3345]

<span>a. To determine the velocity at which the camera strikes the ground: v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera impacts the ground with a speed of 42 m/s. b. To calculate the duration it takes for the camera to reach the bottom: y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds

The camera descends for 11.4 seconds before hitting the ground.</span>

3 0

2 months ago