/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q87P In Fig.10聽-聽55, a wheel of rad... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Q87P (page 293)

In Fig.10-55, a wheel of radius 0.20mis mounted on a frictionless horizontal axle. A massless cord is wrapped around the wheel and attached to a2.0Kgbox that slides on a frictionless surface inclined at angle 胃=20掳with the horizontal. The box accelerates down the surface at2.0ms2. What is the rotational inertia of the wheel about the axle?

Short Answer

Expert verified

Rotational inertia about wheel is0.054kgm2

Step by step solution

01

Given

  1. Radius of wheel is0.2m
  2. Mass of box is2.0kg
  3. Acceleration is2.0m/s2
  4. Angle of inclination is20掳
02

To understand the concept

Use Newton鈥檚 second law to find tension. Using the formula for the torque in terms of tension and radius and also using the formula in terms of moment of inertia and angular acceleration, find the acceleration.

Formula:

Fnet=M脳a蟿=I伪=T脳r伪=ar

03

Draw the Free Body Diagram

04

Calculate the rotational inertia of the wheel about the axle

Net force acting on block is as follow

Fnet=magsin胃-TA

Here胃=20掳

鈬�Fnet=magsin20-TA

According to Newton鈥檚 law,

Fnet=maa

So

maa=magsin20-TA鈬�2脳2=2脳9.8脳sin20-TA鈬�TA=2.7N

Here torque is due to tension so

蟿=r脳T=I脳伪鈬�r脳TA=I脳ar鈬�I=TA脳r2a鈬�I=2.7脳0.222鈬�I=0.054kgm2

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In Fig.10-31 , wheel A of radius rA=10鈥�cmis coupled by belt B to wheel C of radius rC=25鈥�cm .The angular speed of wheel A is increased from rest at a constant rate of1.6rads2 . Find the time needed for wheel C to reach an angular speed of 100revmin , assuming the belt does not slip. (Hint: If the belt does not slip, the linear speeds at the two rims must be equal.)

The uniform solid block in Fig 10-38has mass 0.172kg and edge lengths a = 3.5cm, b = 8.4cm, and c = 1.4cm. Calculate its rotational inertia about an axis through one corner and perpendicular to the large faces.

Calculate the rotational inertia of a meter stick, with mass 0.56kg, about an axis perpendicular to the stick and located at the 20cmmark. (Treat the stick as a thin rod.)

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch? Answer in radians per second.

In Fig10-61., four pulleys are connected by two belts. Pulley A (radius15鈥塩尘) is the drive pulley, and it rotates at.10鈥塺补诲/蝉Pulley B (radius10鈥塩尘) is connected by belt 1to pulley A. Pulley B鈥� (radius5鈥塩尘) is concentric with pulley B and is rigidly attached to it. Pulley C (radius25鈥塩尘) is connected by belt 2to pulley B鈥�. Calculate (a) the linear speed of a point on belt 1, (b) the angularspeed of pulley B, (c) the angular speed of pulley B鈥�, (d) the linear speed of a point on belt2, and (e) the angular speed of pulley C. (Hint: If the belt between two pulleys does not slip, the linear speeds at the rims of the two pulleys must be equal.)
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Modern Physics

Read Explanation

Space Physics

Read Explanation

Electricity

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.